Measures of dispersion

1. Revision of range

ID is: 3151 Seed is: 5565

Working with the range of a data set

Determine the minimum value in a data set if the maximum value in the data set is $\text{43}$ and the range of the data is $\text{0}$ .

Answer: The minimum value is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the range is the difference between the largest and smallest values in a data set.

STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

The range of a data set tells us how spread out the data are: it is the difference between the largest and smallest values. In this question, we do not know what the data values are, but we know that the range of the data is $\text{0}$ . We also know that the maximum value is $\text{43}$ .

We can solve this question using the equation for the range:

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

However, for this solution let's think through the information that we have. In this case, the range is zero. That means there is no spread in the data: all the values must be equal. In fact, the data must be something like this:

\begin{aligned} {43; 43; 43} \end{aligned}

$\begin{aligned} \{43\; ; \;43\; ; \;43\} \end{aligned}$

We do not know how many values there are in the set (there is no information about that). But if the range is zero then there is no separation between the largest and smallest values. All the values must be the same.

The minimum value is $\text{43}$ .

Submit your answer as:

ID is: 3151 Seed is: 8043

Working with the range of a data set

Consider a data set with a maximum value of $\text{22}$ . What is the minimum value in the data set if the range of the data is $\text{0}$ ?

Answer: The minimum value is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the range is the difference between the largest and smallest values in a data set.

STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

We can solve this question using the equation for the range:

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

\begin{aligned} {22; 22; 22; 22} \end{aligned}

$\begin{aligned} \{22\; ; \;22\; ; \;22\; ; \;22\} \end{aligned}$

The minimum value is $\text{22}$ .

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ID is: 3151 Seed is: 2053

Working with the range of a data set

Consider a data set with a maximum value of $\text{33}$ . What is the minimum value in the data set if the range of the data is $\text{17}$ ?

Answer: The minimum value is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the range is the difference between the largest and smallest values in a data set.

STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

The range of a data set tells us how spread out the data are: it is the difference between the largest and smallest values. In this question, we do not know what the data values are, but we know that the range of the data is $\text{17}$ . We also know that the maximum value is $\text{33}$ .

We can solve this question using the equation for the range:

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

However, for this solution let's think through the information that we have. The maximum value in the data set is $\text{33}$ and the range is $\text{17}$ . So the minimum value is $\text{17}$ below $\text{33}$ .

\begin{aligned} minimum & = maximum - range \\ = 33 - 17 \\ = 16 \end{aligned}

$\begin{aligned} \text{minimum } & = \text{ maximum - range}\\ \normalsize & = \text{33} - \text{17}\\ & = \text{16} \end{aligned}$

The minimum value is $\text{16}$ .

Submit your answer as:

ID is: 3154 Seed is: 1391

Finding the range by solving an equation

Here is some information about a certain data set:

The minimum value is $- 6 a$ .
The maximum value is $9$ .
The range of the data is equal to $- 3 \left(a - 5\right)$ .

Use this information to determine the range of data set.

Answer: The range of the data is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for $a$ .

STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of $a$ . We can solve this question using the definition of the range.

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

Substituting in the expressions from the question, we get:

\begin{matrix} - 3 (a - 5) = (9) - (- 6 a) \end{matrix}

$\begin{array}{c}- 3 \left(a - 5\right) = \left( 9 \right) - \left( - 6 a \right)\end{array}$

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Start by expanding the expression on the left.There is also a double negative on the right hand side changes to addition.

\begin{aligned} - 3 a + 15 & = 9 + 6 a \\ - 3 a + 15 & = 6 a + 9 \\ - 9 a & = - 6 \\ a & = \frac{2}{3} \end{aligned}

$\begin{aligned} - 3 a + 15 & = 9 + 6 a \\ - 3 a + 15 & = 6 a + 9 \\ - 9 a & = -6 \\ a & = \frac{2}{3} \end{aligned}$

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that $a = \frac{2}{3}$ . But we are not finished yet. The question asked us to find the range of the data set. We can use the value of $a$ to do that. Just substitute $a = \frac{2}{3}$ into the expression for the range. Use the expanded expression for the range, which we found above (this saves a little time).

\begin{aligned} range & = - 3 a + 15 \\ = - 3 (\frac{2}{3}) + 15 \\ = 13 \end{aligned}

$\begin{aligned} \small \text{range} & = - 3 a + 15 \\ & = - 3 \left( \frac{2}{3} \right) + 15 \\ & = 13 \end{aligned}$

The range of the data set is $\text{13}$ .

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ID is: 3154 Seed is: 3712

Finding the range by solving an equation

Here is some information about a certain data set:

The minimum value is $- 6 n + 17$ .
The maximum value is $34$ .
The range of the data is equal to $- 8 \left(n - 3\right)$ .

Use this information to determine the range of data set.

Answer: The range of the data is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for $n$ .

STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of $n$ . We can solve this question using the definition of the range.

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

Substituting in the expressions from the question, we get:

\begin{matrix} - 8 (n - 3) = (34) - (- 6 n + 17) \end{matrix}

$\begin{array}{c}- 8 \left(n - 3\right) = \left( 34 \right) - \left( - 6 n + 17 \right)\end{array}$

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Start by expanding the expression on the left. Notice that on the right hand side we also must distribute the negative sign into the binomial $- 6 n + 17$ .

\begin{aligned} - 8 n + 24 & = 34 - (- 6 n + 17) \\ - 8 n + 24 & = 34 + 6 n - 17 \\ - 8 n + 24 & = 6 n + 17 \\ - 14 n & = - 7 \\ n & = \frac{1}{2} \end{aligned}

$\begin{aligned} - 8 n + 24 & = 34 - \left( - 6 n + 17 \right)\\ - 8 n + 24 & = 34 + 6 n - 17 \\ - 8 n + 24 & = 6 n + 17 \\ - 14 n & = -7 \\ n & = \frac{1}{2} \end{aligned}$

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that $n = \frac{1}{2}$ . But we are not finished yet. The question asked us to find the range of the data set. We can use the value of $n$ to do that. Just substitute $n = \frac{1}{2}$ into the expression for the range. Use the expanded expression for the range, which we found above (this saves a little time).

\begin{aligned} range & = - 8 n + 24 \\ = - 8 (\frac{1}{2}) + 24 \\ = 20 \end{aligned}

$\begin{aligned} \small \text{range} & = - 8 n + 24 \\ & = - 8 \left( \frac{1}{2} \right) + 24 \\ & = 20 \end{aligned}$

The range of the data set is $\text{20}$ .

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ID is: 3154 Seed is: 4141

Finding the range by solving an equation

Here is some information about a certain data set:

The minimum value is $- 3 a - 15$ .
The maximum value is $9 a - 26$ .
The range of the data is equal to $9 a - 7$ .

Use this information to determine the range of data set.

Answer: The range of the data is: .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for $a$ .

STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of $a$ . We can solve this question using the definition of the range.

\begin{matrix} range = maximum - minimum \end{matrix}

$\begin{array}{c}\text{range = maximum - minimum}\end{array}$

Substituting in the expressions from the question, we get:

\begin{matrix} 9 a - 7 = (9 a - 26) - (- 3 a - 15) \end{matrix}

$\begin{array}{c}9 a - 7 = \left( 9 a - 26 \right) - \left( - 3 a - 15 \right)\end{array}$

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Notice that on the right hand side we must distribute the negative sign into the binomial $- 3 a - 15$ .

\begin{aligned} 9 a - 7 & = 9 a - 26 - (- 3 a - 15) \\ 9 a - 7 & = 9 a - 26 + 3 a + 15 \\ 9 a - 7 & = 12 a - 11 \\ - 3 a & = - 4 \\ a & = \frac{4}{3} \end{aligned}

$\begin{aligned} 9 a - 7 & = 9 a - 26 - \left( - 3 a - 15 \right)\\ 9 a - 7 & = 9 a - 26 + 3 a + 15 \\ 9 a - 7 & = 12 a - 11 \\ - 3 a & = -4 \\ a & = \frac{4}{3} \end{aligned}$

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that $a = \frac{4}{3}$ . But we are not finished yet. The question asked us to find the range of the data set. We can use the value of $a$ to do that. Just substitute $a = \frac{4}{3}$ into the expression for the range.

\begin{aligned} range & = 9 a - 7 \\ = 9 (\frac{4}{3}) - 7 \\ = 5 \end{aligned}

$\begin{aligned} \small \text{range} & = 9 a - 7 \\ & = 9 \left( \frac{4}{3} \right) - 7 \\ & = 5 \end{aligned}$

The range of the data set is $\text{5}$ .

Submit your answer as:

ID is: 4351 Seed is: 9011

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of devices in use by $\text{19}$ schools for Siyavula Practice.

\begin{array}{ccccc} 119 & 98 & 90 & 155 & 181 \\ 151 & 68 & 134 & 171 & 107 \\ 95 & 61 & 131 & 80 & 96 \\ 144 & 101 & 55 & 117 \end{array}

$\begin{array}{ccccc} 119&98&90&155&181\\151&68&134&171&107\\95&61&131&80&96\\144&101&55&117 \end{array}$

Determine the median and the interquartile range of the data.
Answer:
- The median is $\text{devices}$ .
- The interquartile range is $\text{devices}$ .
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
- The median is the middle value in an ordered list.
- The upper quartile is the median of the upper half of the data values.
- The lower quartile is the median of the lower half of the data values.
- The interquartile range is the difference between the upper and lower quartiles.
STEP: Determine the median of the data set
[−1 point ⇒ 2 / 3 points left]

The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

The data values:

$\begin{array}{ccccc} 119&98&90&155&181\\151&68&134&171&107\\95&61&131&80&96\\144&101&55&117 \end{array}$

The ordered data values:

$\begin{array}{ccccc} 55&61&68&80&90\\95&96&98&101 \\ \\ & & { \color{red}{ \textbf{107} } } \\ \\ 117&119&131&134&144\\151&155&171&181 \end{array}$

The data set has $\text{19}$ data values. The median is the tenth value in the ordered list, which is $\text{107}\ \text{devices}$ .

STEP: Determine the interquartile range
[−2 points ⇒ 0 / 3 points left]
- The median is the ‘middle value’ of the whole data set, referred to as $Q_2$ .
- The lower quartile is the median of the lower half of the data values, referred to as $Q_1$ .
- The upper quartile is the median of the upper half of the data values, referred to as $Q_3$ .
- The interquartile range is the difference between the upper and lower quartiles, referred to as $IQR$ . The formula is: $IQR = Q_3 - Q_1$ .
We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

The ordered data values:

$\begin{array}{ccccc} 55&61&68&80&{ \color{blue}{ \textbf{90} } }\\95&96&98&101 \\ \\ & & { \color{red}{ \textbf{107} } } \\ \\ 117&119&131&134&{ \color{purple}{ \textbf{144} } }\\151&155&171&181 \end{array}$

Calculate the lower and upper quartiles:

$\begin{aligned} {\color{blue}{Q_1}} &= T_{5} = \text{90}\ \text{devices} \end{aligned}$

$\begin{aligned} {\color{purple}{Q_3}} &= T_{15} = \text{144}\ \text{devices} \end{aligned}$

Calculate the interquartile range:

$\begin{aligned} IQR &= {\color{purple}{Q_3}} - {\color{blue}{Q_1}} \\ &= \text{144} - \text{90} \\ &= \text{54}\ \text{devices} \end{aligned}$

Submit your answer as: and
On a piece of paper, draw a box and whisker diagram for the given data set:

$\begin{array}{ccccc} 55&61&68&80&90\\95&96&98&101&107\\117&119&131&134&144\\151&155&171&181 \end{array}$

NOTE: You should always draw a box and whisker diagram on a scaled number line.

Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)
Answer:
The value at:
- $a$ is $\text{devices}$ .
- $b$ is $\text{devices}$ .
- $c$ is $\text{devices}$ .
- $d$ is $\text{devices}$ .
- $e$ is $\text{devices}$ .
numeric
one-of
type(numeric.abserror(0.01))
numeric
one-of
type(numeric.abserror(0.01))
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

When you draw a box and whisker diagram, you mark the following values on the number line:
- The minimum value
- The lower quartile
- The median of the whole data set
- The upper quartile
- The maximum value
STEP: Draw box and whisker diagram
[−3 points ⇒ 0 / 3 points left]
A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:
- A number line is drawn showing the range of the data set.
- A box, representing the interquartile range, is drawn on the number line.
- The left edge of the box shows the lower quartile.
- The right edge of the box shows the upper quartile.
- A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
- The median of the whole data set is shown as a vertical line inside the box.
The labels you need for your box and whisker diagram are:
- $a$ shows the lowest value in the data set, the minimum value, which is $\text{55}\ \text{devices}$ .
- $b$ shows the left edge of the box, the lower quartile of the data set, which is $\color{blue}{\text{90}}$ $\text{devices}$ .
- $c$ shows the line inside the box, the median of the whole data set, which is $\color{red}{\text{107}}$ $\text{devices}$ .
- $d$ shows the right edge of the box, the upper quartile of the data set, which is $\color{purple}{\text{144}}$ $\text{devices}$ .
- $e$ shows the highest value in the data set, the maximum value, which is $\text{181}\ \text{devices}$ .
When drawn to scale, the box and whisker diagram looks like this:

On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels $Q_1$ , $Q_2$ , and $Q_3$ .

Submit your answer as: andandandand

ID is: 4351 Seed is: 4795

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of books donated by $\text{11}$ classes for a literacy programme.

\begin{array}{ccccc} 74 & 37 & 40 & 34 & 5 \\ 56 & 55 & 20 & 31 & 71 \\ 42 \end{array}

$\begin{array}{ccccc} 74&37&40&34&5\\56&55&20&31&71\\42 \end{array}$

Determine the median and the interquartile range of the data.
Answer:
- The median is $\text{books}$ .
- The interquartile range is $\text{books}$ .
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
- The median is the middle value in an ordered list.
- The upper quartile is the median of the upper half of the data values.
- The lower quartile is the median of the lower half of the data values.
- The interquartile range is the difference between the upper and lower quartiles.
STEP: Determine the median of the data set
[−1 point ⇒ 2 / 3 points left]

The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

The data values:

$\begin{array}{ccccc} 74&37&40&34&5\\56&55&20&31&71\\42 \end{array}$

The ordered data values:

$\begin{array}{ccccc} 5&20&31&34&37 \\ \\ & & { \color{red}{ \textbf{40} } } \\ \\ 42&55&56&71&74 \end{array}$

The data set has $\text{11}$ data values. The median is the sixth value in the ordered list, which is $\text{40}\ \text{books}$ .

STEP: Determine the interquartile range
[−2 points ⇒ 0 / 3 points left]
- The median is the ‘middle value’ of the whole data set, referred to as $Q_2$ .
- The lower quartile is the median of the lower half of the data values, referred to as $Q_1$ .
- The upper quartile is the median of the upper half of the data values, referred to as $Q_3$ .
- The interquartile range is the difference between the upper and lower quartiles, referred to as $IQR$ . The formula is: $IQR = Q_3 - Q_1$ .
We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

The ordered data values:

$\begin{array}{ccccc} 5&20&{ \color{blue}{ \textbf{31} } }&34&37 \\ \\ & & { \color{red}{ \textbf{40} } } \\ \\ 42&55&{ \color{purple}{ \textbf{56} } }&71&74 \end{array}$

Calculate the lower and upper quartiles:

$\begin{aligned} {\color{blue}{Q_1}} &= T_{3} = \text{31}\ \text{books} \end{aligned}$

$\begin{aligned} {\color{purple}{Q_3}} &= T_{9} = \text{56}\ \text{books} \end{aligned}$

Calculate the interquartile range:

$\begin{aligned} IQR &= {\color{purple}{Q_3}} - {\color{blue}{Q_1}} \\ &= \text{56} - \text{31} \\ &= \text{25}\ \text{books} \end{aligned}$

Submit your answer as: and
On a piece of paper, draw a box and whisker diagram for the given data set:

$\begin{array}{ccccc} 5&20&31&34&37\\40&42&55&56&71\\74 \end{array}$

NOTE: You should always draw a box and whisker diagram on a scaled number line.

Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)
Answer:
The value at:
- $a$ is $\text{books}$ .
- $b$ is $\text{books}$ .
- $c$ is $\text{books}$ .
- $d$ is $\text{books}$ .
- $e$ is $\text{books}$ .
numeric
one-of
type(numeric.abserror(0.01))
numeric
one-of
type(numeric.abserror(0.01))
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

When you draw a box and whisker diagram, you mark the following values on the number line:
- The minimum value
- The lower quartile
- The median of the whole data set
- The upper quartile
- The maximum value
STEP: Draw box and whisker diagram
[−3 points ⇒ 0 / 3 points left]
A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:
- A number line is drawn showing the range of the data set.
- A box, representing the interquartile range, is drawn on the number line.
- The left edge of the box shows the lower quartile.
- The right edge of the box shows the upper quartile.
- A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
- The median of the whole data set is shown as a vertical line inside the box.
The labels you need for your box and whisker diagram are:
- $a$ shows the lowest value in the data set, the minimum value, which is $\text{5}\ \text{books}$ .
- $b$ shows the left edge of the box, the lower quartile of the data set, which is $\color{blue}{\text{31}}$ $\text{books}$ .
- $c$ shows the line inside the box, the median of the whole data set, which is $\color{red}{\text{40}}$ $\text{books}$ .
- $d$ shows the right edge of the box, the upper quartile of the data set, which is $\color{purple}{\text{56}}$ $\text{books}$ .
- $e$ shows the highest value in the data set, the maximum value, which is $\text{74}\ \text{books}$ .
When drawn to scale, the box and whisker diagram looks like this:

On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels $Q_1$ , $Q_2$ , and $Q_3$ .

Submit your answer as: andandandand

ID is: 4351 Seed is: 7970

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of ecobricks made by $\text{25}$ classes during Enviro Week.

\begin{array}{ccccc} 69 & 23 & 55 & 10 & 31 \\ 8 & 64 & 9 & 48 & 38 \\ 18 & 62 & 45 & 65 & 32 \\ 17 & 70 & 43 & 47 & 35 \\ 39 & 40 & 28 & 53 & 5 \end{array}

$\begin{array}{ccccc} 69&23&55&10&31\\8&64&9&48&38\\18&62&45&65&32\\17&70&43&47&35\\39&40&28&53&5 \end{array}$

Determine the median and the interquartile range of the data.
Answer:
- The median is $\text{ecobricks}$ .
- The interquartile range is $\text{ecobricks}$ .
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
- The median is the middle value in an ordered list.
- The upper quartile is the median of the upper half of the data values.
- The lower quartile is the median of the lower half of the data values.
- The interquartile range is the difference between the upper and lower quartiles.
STEP: Determine the median of the data set
[−1 point ⇒ 2 / 3 points left]

The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

The data values:

$\begin{array}{ccccc} 69&23&55&10&31\\8&64&9&48&38\\18&62&45&65&32\\17&70&43&47&35\\39&40&28&53&5 \end{array}$

The ordered data values:

$\begin{array}{ccccc} 5&8&9&10&17\\18&23&28&31&32\\35&38 \\ \\ & & { \color{red}{ \textbf{39} } } \\ \\ 40&43&45&47&48\\53&55&62&64&65\\69&70 \end{array}$

The data set has $\text{25}$ data values. The median is the $13^{\text{th}}$ value in the ordered list, which is $\text{39}\ \text{ecobricks}$ .

STEP: Determine the interquartile range
[−2 points ⇒ 0 / 3 points left]
- The median is the ‘middle value’ of the whole data set, referred to as $Q_2$ .
- The lower quartile is the median of the lower half of the data values, referred to as $Q_1$ .
- The upper quartile is the median of the upper half of the data values, referred to as $Q_3$ .
- The interquartile range is the difference between the upper and lower quartiles, referred to as $IQR$ . The formula is: $IQR = Q_3 - Q_1$ .
We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

The ordered data values:

$\begin{array}{ccccc} 5&8&9&10&17\\{ \color{blue}{ \textbf{18} } }&{ \color{blue}{ \textbf{23} } }&28&31&32\\35&38 \\ \\ & & { \color{red}{ \textbf{39} } } \\ \\ 40&43&45&47&48\\{ \color{purple}{ \textbf{53} } }&{ \color{purple}{ \textbf{55} } }&62&64&65\\69&70 \end{array}$

Calculate the lower and upper quartiles:

$\begin{aligned} {\color{blue}{Q_1}} &= \dfrac{ T_{6} + T_{7} }{ 2 } \\&= \dfrac{ 18 + 23 }{ 2 } \\&= \text{20.5} \end{aligned}$

$\begin{aligned} {\color{purple}{Q_3}} &= \dfrac{ T_{19} + T_{20} }{ 2 } \\&= \dfrac{ 53 + 55 }{ 2 } \\&= \text{54} \end{aligned}$

Calculate the interquartile range:

$\begin{aligned} IQR &= {\color{purple}{Q_3}} - {\color{blue}{Q_1}} \\ &= \text{54} - \text{20.5} \\ &= \text{33.5}\ \text{ecobricks} \end{aligned}$

Submit your answer as: and
On a piece of paper, draw a box and whisker diagram for the given data set:

$\begin{array}{ccccc} 5&8&9&10&17\\18&23&28&31&32\\35&38&39&40&43\\45&47&48&53&55\\62&64&65&69&70 \end{array}$

NOTE: You should always draw a box and whisker diagram on a scaled number line.

Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)
Answer:
The value at:
- $a$ is $\text{ecobricks}$ .
- $b$ is $\text{ecobricks}$ .
- $c$ is $\text{ecobricks}$ .
- $d$ is $\text{ecobricks}$ .
- $e$ is $\text{ecobricks}$ .
numeric
one-of
type(numeric.abserror(0.01))
numeric
one-of
type(numeric.abserror(0.01))
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

When you draw a box and whisker diagram, you mark the following values on the number line:
- The minimum value
- The lower quartile
- The median of the whole data set
- The upper quartile
- The maximum value
STEP: Draw box and whisker diagram
[−3 points ⇒ 0 / 3 points left]
A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:
- A number line is drawn showing the range of the data set.
- A box, representing the interquartile range, is drawn on the number line.
- The left edge of the box shows the lower quartile.
- The right edge of the box shows the upper quartile.
- A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
- The median of the whole data set is shown as a vertical line inside the box.
The labels you need for your box and whisker diagram are:
- $a$ shows the lowest value in the data set, the minimum value, which is $\text{5}\ \text{ecobricks}$ .
- $b$ shows the left edge of the box, the lower quartile of the data set, which is $\color{blue}{\text{20.5}}$ $\text{ecobricks}$ .
- $c$ shows the line inside the box, the median of the whole data set, which is $\color{red}{\text{39}}$ $\text{ecobricks}$ .
- $d$ shows the right edge of the box, the upper quartile of the data set, which is $\color{purple}{\text{54}}$ $\text{ecobricks}$ .
- $e$ shows the highest value in the data set, the maximum value, which is $\text{70}\ \text{ecobricks}$ .
When drawn to scale, the box and whisker diagram looks like this:

On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels $Q_1$ , $Q_2$ , and $Q_3$ .

Submit your answer as: andandandand

ID is: 3153 Seed is: 7597

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the $25^{\text{th}}$ percentile.

Answer: The

25^{th}

$25^{\text{th}}$ percentile corresponds to the .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the

50^{th}

$50^{\text{th}}$ percentile is halfway from the minimum value to the maximum value.

STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile	Position in the data	Value in the data set
$0^{\text{th}}$	beginning	minimum value
$25^{\text{th}}$	1 quarter of the way	first quartile
$50^{\text{th}}$	2 quarters of the way	median (second quartile)
$75^{\text{th}}$	3 quarters of the way	third quartile
$100^{\text{th}}$	upper limit of the data	maximum value

For this question we have the $25^{\text{th}}$ percentile. This corresponds to the first quartile.

The correct choice from the dropdown menu is: first quartile.

Submit your answer as:

ID is: 3153 Seed is: 1673

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the $25^{\text{th}}$ percentile.

Answer: The

25^{th}

$25^{\text{th}}$ percentile corresponds to the .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the

50^{th}

$50^{\text{th}}$ percentile is halfway from the minimum value to the maximum value.

STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile	Position in the data	Value in the data set
$0^{\text{th}}$	beginning	minimum value
$25^{\text{th}}$	1 quarter of the way	first quartile
$50^{\text{th}}$	2 quarters of the way	median (second quartile)
$75^{\text{th}}$	3 quarters of the way	third quartile
$100^{\text{th}}$	upper limit of the data	maximum value

For this question we have the $25^{\text{th}}$ percentile. This corresponds to the first quartile.

The correct choice from the dropdown menu is: first quartile.

Submit your answer as:

ID is: 3153 Seed is: 6233

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the $17^{\text{th}}$ percentile.

Answer: The

17^{th}

$17^{\text{th}}$ percentile corresponds to the .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the

50^{th}

$50^{\text{th}}$ percentile is halfway from the minimum value to the maximum value.

STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile	Position in the data	Value in the data set
$0^{\text{th}}$	beginning	minimum value
$25^{\text{th}}$	1 quarter of the way	first quartile
$50^{\text{th}}$	2 quarters of the way	median (second quartile)
$75^{\text{th}}$	3 quarters of the way	third quartile
$100^{\text{th}}$	upper limit of the data	maximum value

For this question we have the $17^{\text{th}}$ percentile. This does not correspond to any of the quartile boundaries, which are all multiples of 25.

The correct choice from the dropdown menu is: none of the above.

Submit your answer as:

ID is: 3185 Seed is: 3455

Percentile and rank in a data set

The data set below shows a total of 14 values.

\begin{matrix} {a; c; g; i; j; k; l; m; o; r; s; u; v; w} \end{matrix}

$\begin{array}{c}\{ a; \, c; \, g; \, i; \, j; \, k; \, l; \, m; \, o; \, r; \, s; \, u; \, v; \, w \}\end{array}$

Note that the data are in alphabetical order.

Find the rank of the data value at the tenth percentile.

INSTRUCTION: Do not round off your answer.
Answer: The rank of the tenth percentile is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the following equation to convert the percentile given in the question to a rank:

$\begin{array}{c}r = \dfrac{p}{100}(n - 1) + 1\end{array}$

STEP: Identify the values needed for the solution
[−1 point ⇒ 1 / 2 points left]

We have a data set with 14 items in it, and the question asks us to determine the rank of the tenth percentile in the data set. For this we use the formula:

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:
- $p$ is the percentile value
- $n$ is the number of values in the data set
- $r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set
In this case, we have the values:

$\begin{aligned} p & = \text{10}\\ n & = \text{14} \end{aligned}$

STEP: Substitute into the formula and evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now substitute the values $p$ and $n$ into the formula and evaluate to determine $r$ . As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(\text{10})}{\text{100}} \times \left((14) - 1\right) + 1\\ & = \frac{10}{100} \times (13) + 1\\ & = \text{1.3} + 1\\ & = \text{2.3} \end{aligned}$

The rank of the tenth percentile in the data set is $\text{2.3}$ .

Submit your answer as:
Identify the value in the data set which is at or closest to the tenth percentile. If there is more than one data value closest to the tenth percentile, give any one answer.

Answer: The value at or closest to the tenth percentile is: .
one-of
type(string)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the rank value from Question 1 to find the value in the data set closest to the tenth percentile.

STEP: Use the rank of the tenth percentile to find the correct value in the data set
[−1 point ⇒ 0 / 1 points left]

Now we need to find the value in the data set which corresponds to the tenth percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of $\text{2.3}$ .

In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to $\text{2.3}$ .

The rank value $\text{2.3}$ is closest to second position in the data set. So the tenth percentile corresponds to $c$ .

The value at or closest to the tenth percentile is $c$ .

Submit your answer as:

ID is: 3185 Seed is: 6814

Percentile and rank in a data set

The data set below shows a total of 7 values.

\begin{matrix} {- 20; - 16; - 8; - 3; 2; 14; 25} \end{matrix}

$\begin{array}{c}\{ -20; \, -16; \, -8; \, -3; \, 2; \, 14; \, 25 \}\end{array}$

Note that the data are in numeric order.

Find the rank of the data value at the $75^{\text{th}}$ percentile.

INSTRUCTION: Do not round off your answer.
Answer: The rank of the $75^{\text{th}}$ percentile is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the following equation to convert the percentile given in the question to a rank:

$\begin{array}{c}r = \dfrac{p}{100}(n - 1) + 1\end{array}$

STEP: Identify the values needed for the solution
[−1 point ⇒ 1 / 2 points left]

We have a data set with 7 items in it, and the question asks us to determine the rank of the $75^{\text{th}}$ percentile in the data set. For this we use the formula:

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:
- $p$ is the percentile value
- $n$ is the number of values in the data set
- $r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set
In this case, we have the values:

$\begin{aligned} p & = \text{75}\\ n & = \text{7} \end{aligned}$

STEP: Substitute into the formula and evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now substitute the values $p$ and $n$ into the formula and evaluate to determine $r$ . As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(\text{75})}{\text{100}} \times \left((7) - 1\right) + 1\\ & = \frac{75}{100} \times (6) + 1\\ & = \text{4.5} + 1\\ & = \text{5.5} \end{aligned}$

The rank of the $75^{\text{th}}$ percentile in the data set is $\text{5.5}$ .

Submit your answer as:
Figure out the value in the data set which is at or closest to the $75^{\text{th}}$ percentile. If there is more than one data value closest to the $75^{\text{th}}$ percentile, give any one answer.

Answer: The value at or closest to the $75^{\text{th}}$ percentile is: .
one-of
type(numeric.noerror)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the rank value from Question 1 to find the value in the data set closest to the $75^{\text{th}}$ percentile.

STEP: Use the rank of the $75^{\text{th}}$ percentile to find the correct value in the data set
[−1 point ⇒ 0 / 1 points left]

Now we need to find the value in the data set which corresponds to the $75^{\text{th}}$ percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of $\text{5.5}$ .

In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to $\text{5.5}$ .

The rank value $\text{5.5}$ is exactly between the fifth and sixth positions. So the $75^{\text{th}}$ percentile corresponds equally well to both $2$ and other $14$ .

Either of these answers is acceptable for the value closest to the $75^{\text{th}}$ percentile: $2$ or $14$ .

Submit your answer as:

ID is: 3185 Seed is: 9389

Percentile and rank in a data set

The data set below shows a total of 12 values.

\begin{matrix} {- 26; - 21; - 20; - 8; - 5; - 1; 6; 7; 8; 9; 13; 18} \end{matrix}

$\begin{array}{c}\{ -26; \, -21; \, -20; \, -8; \, -5; \, -1; \, 6; \, 7; \, 8; \, 9; \, 13; \, 18 \}\end{array}$

Note that the data are in numeric order.

Find the rank of the data value at the $49^{\text{th}}$ percentile.

INSTRUCTION: Do not round off your answer.
Answer: The rank of the $49^{\text{th}}$ percentile is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the following equation to convert the percentile given in the question to a rank:

$\begin{array}{c}r = \dfrac{p}{100}(n - 1) + 1\end{array}$

STEP: Identify the values needed for the solution
[−1 point ⇒ 1 / 2 points left]

We have a data set with 12 items in it, and the question asks us to determine the rank of the $49^{\text{th}}$ percentile in the data set. For this we use the formula:

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:
- $p$ is the percentile value
- $n$ is the number of values in the data set
- $r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set
In this case, we have the values:

$\begin{aligned} p & = \text{49}\\ n & = \text{12} \end{aligned}$

STEP: Substitute into the formula and evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now substitute the values $p$ and $n$ into the formula and evaluate to determine $r$ . As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(\text{49})}{\text{100}} \times \left((12) - 1\right) + 1\\ & = \frac{49}{100} \times (11) + 1\\ & = \text{5.39} + 1\\ & = \text{6.39} \end{aligned}$

The rank of the $49^{\text{th}}$ percentile in the data set is $\text{6.39}$ .

Submit your answer as:
Determine the value in the data set which is at or closest to the $49^{\text{th}}$ percentile. If there is more than one data value closest to the $49^{\text{th}}$ percentile, give any one answer.

Answer: The value at or closest to the $49^{\text{th}}$ percentile is: .
one-of
type(numeric.noerror)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the rank value from Question 1 to find the value in the data set closest to the $49^{\text{th}}$ percentile.

STEP: Use the rank of the $49^{\text{th}}$ percentile to find the correct value in the data set
[−1 point ⇒ 0 / 1 points left]

Now we need to find the value in the data set which corresponds to the $49^{\text{th}}$ percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of $\text{6.39}$ .

In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to $\text{6.39}$ .

The rank value $\text{6.39}$ is closest to sixth position in the data set. So the $49^{\text{th}}$ percentile corresponds to $-1$ .

The value at or closest to the $49^{\text{th}}$ percentile is $-1$ .

Submit your answer as:

ID is: 3186 Seed is: 9097

Applying percentile to a data set

The data set below has already been ordered for you. Find the item which corresponds to the $62^{\text{nd}}$ percentile.

\begin{matrix} {- 28; - 11; - 9; - 7; - 5; 6; 14; 15; 18; 19; 21; 21} \end{matrix}

$\begin{array}{c}\{ -28; \, -11; \, -9; \, -7; \, -5; \, 6; \, 14; \, 15; \, 18; \, 19; \, 21; \, 21 \}\end{array}$

Answer: The item at or closest to the

62^{nd}

$62^{\text{nd}}$ percentile is: .

one-of

type(numeric.noerror)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the rank of the item at the $62^{\text{nd}}$ percentile.

STEP: Determine the rank of the item at the

62^{nd}

$62^{\text{nd}}$ percentile
[−1 point ⇒ 1 / 2 points left]

We have a data set with $\text{12}$ items in it, and the question asks us to determine the rank of the $62^{\text{nd}}$ percentile in the data set. For this we use the formula:

\begin{matrix} r = \frac{p}{100} (n - 1) + 1 \end{matrix}

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:

$p$ is the percentile value
$n$ is the number of items in the data set
$r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set

For this question, we have $p = 62$ and $n = 12$ . Substitute these values into the formula above to find the rank.

\begin{aligned} r & = \frac{p}{100} (n - 1) + 1 \\ = \frac{(62)}{100} \times ((12) - 1) + 1 \\ = \frac{62}{100} \times (11) + 1 \\ = 6.82 + 1 \\ = 7.82 \end{aligned}

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(62)}{100} \times \left((12) - 1\right) + 1\\ & = \frac{62}{100} \times (11) + 1\\ & = \text{6.82} + 1\\ & = \text{7.82} \end{aligned}$

The rank of the $62^{\text{nd}}$ percentile in the data set is $\text{7.82}$ .

STEP: Use the rank of the

62^{nd}

$62^{\text{nd}}$ percentile to find the correct item in the data set
[−1 point ⇒ 0 / 2 points left]

Now we need to find the item in the data set which corresponds to the $62^{\text{nd}}$ percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of $\text{7.82}$ .

But the rank of an item must be an integer because it represents a position. We need to find the closest position value to $\text{7.82}$ .

The rank value $\text{7.82}$ is closest to eighth position in the data set. So the $62^{\text{nd}}$ percentile corresponds to $15$ .

The item at the $62^{\text{nd}}$ percentile is $15$ .

Submit your answer as:

ID is: 3186 Seed is: 5396

Applying percentile to a data set

The data set below has already been ordered for you. Which item is at the $0^{\text{th}}$ percentile?

\begin{matrix} {c; h; h; h; j; j; k; l; m; p; q; r; u; x; z} \end{matrix}

$\begin{array}{c}\{ c; \, h; \, h; \, h; \, j; \, j; \, k; \, l; \, m; \, p; \, q; \, r; \, u; \, x; \, z \}\end{array}$

Answer: The item at or closest to the

0^{th}

$0^{\text{th}}$ percentile is: .

one-of

type(string)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You do not need to do any calculations to get this answer: just think about the meaning of the $0^{\text{th}}$ percentile.

STEP: Determine the rank of the item at the

0^{th}

$0^{\text{th}}$ percentile
[−0 points ⇒ 1 / 1 points left]

For this question, it is not necessary to do any calculation in order to determine the item that is at the $0^{\text{th}}$ percentile. The item at the $0^{\text{th}}$ percentile is always the first one in an ordered data set. However we will carry out the working in order to confirm this fact.

We have a data set with $\text{15}$ items in it, and the question asks us to determine the rank of the $0^{\text{th}}$ percentile in the data set. For this we use the formula:

\begin{matrix} r = \frac{p}{100} (n - 1) + 1 \end{matrix}

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:

$p$ is the percentile value
$n$ is the number of items in the data set
$r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set

For this question, we have $p = 0$ and $n = 15$ . Substitute these values into the formula above to find the rank.

\begin{aligned} r & = \frac{p}{100} (n - 1) + 1 \\ = \frac{(0)}{100} \times ((15) - 1) + 1 \\ = \frac{0}{100} \times (14) + 1 \\ = 0 + 1 \\ = 1 \end{aligned}

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(0)}{100} \times \left((15) - 1\right) + 1\\ & = \frac{0}{100} \times (14) + 1\\ & = \text{0} + 1\\ & = \text{1} \end{aligned}$

The rank of the $0^{\text{th}}$ percentile in the data set is $\text{1}$ .

STEP: Use the rank of the

0^{th}

$0^{\text{th}}$ percentile to find the correct item in the data set
[−1 point ⇒ 0 / 1 points left]

Now we need to find the item in the data set which corresponds to the $0^{\text{th}}$ percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of $\text{1}$ .

In this case, we want the first letter in the data set.

The item at or closest to the $0^{\text{th}}$ percentile is $c$ .

Submit your answer as:

ID is: 3186 Seed is: 8637

Applying percentile to a data set

The data set below has already been ordered for you. Determine the item which corresponds to the $85^{\text{th}}$ percentile.

\begin{matrix} {- 26; - 25; - 22; - 18; - 13; 0; 7; 15; 17; 21; 26} \end{matrix}

$\begin{array}{c}\{ -26; \, -25; \, -22; \, -18; \, -13; \, 0; \, 7; \, 15; \, 17; \, 21; \, 26 \}\end{array}$

Answer: The item at or closest to the

85^{th}

$85^{\text{th}}$ percentile is: .

one-of

type(numeric.noerror)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the rank of the item at the $50^{\text{th}}$ percentile.

STEP: Determine the rank of the item at the

85^{th}

$85^{\text{th}}$ percentile
[−1 point ⇒ 1 / 2 points left]

We have a data set with $\text{11}$ items in it, and the question asks us to determine the rank of the $85^{\text{th}}$ percentile in the data set. For this we use the formula:

\begin{matrix} r = \frac{p}{100} (n - 1) + 1 \end{matrix}

$\begin{array}{c}r = \frac{p}{100}(n - 1) + 1\end{array}$

where:

$p$ is the percentile value
$n$ is the number of items in the data set
$r$ is the rank (position) of the $p^{\text{th}}$ percentile in the data set

For this question, we have $p = 85$ and $n = 11$ . Substitute these values into the formula above to find the rank.

\begin{aligned} r & = \frac{p}{100} (n - 1) + 1 \\ = \frac{(85)}{100} \times ((11) - 1) + 1 \\ = \frac{85}{100} \times (10) + 1 \\ = 8.5 + 1 \\ = 9.5 \end{aligned}

$\begin{aligned} r & = \frac{p}{100}(n - 1) + 1\\ & = \frac{(85)}{100} \times \left((11) - 1\right) + 1\\ & = \frac{85}{100} \times (10) + 1\\ & = \text{8.5} + 1\\ & = \text{9.5} \end{aligned}$

The rank of the $85^{\text{th}}$ percentile in the data set is $\text{9.5}$ .

STEP: Use the rank of the

85^{th}

$85^{\text{th}}$ percentile to find the correct item in the data set
[−1 point ⇒ 0 / 2 points left]

Now we need to find the item in the data set which corresponds to the $85^{\text{th}}$ percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of $\text{9.5}$ .

But the rank of an item must be an integer because it represents a position. We need to find the closest position value to $\text{9.5}$ .

The rank value $\text{9.5}$ is exactly between the ninth and tenth positions. So the $85^{\text{th}}$ percentile corresponds equally well to both $17$ and other $21$ .

Either of these answers is acceptable for the item closest to the $85^{\text{th}}$ percentile: $17$ or $21$ .

Submit your answer as:

ID is: 2863 Seed is: 8787

The range

Consider the set of raw data:

\begin{matrix} {5; 9; 10; 11; 7; 10; 5; 10; 7; 6; 9} \end{matrix}

$\begin{array}{c}\{5; 9; 10; 11; 7; 10; 5; 10; 7; 6; 9\}\end{array}$

What is the range of the data set?

Answer: Range:

numeric

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

\begin{matrix} {5; 5; 6; 7; 7; 9; 9; 10; 10; 10; 11} \end{matrix}

$\begin{array}{c}\{5; 5; 6; 7; 7; 9; 9; 10; 10; 10; 11\}\end{array}$

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

Organise the data
Identify the lowest value
Identify the highest value and subtract them

\begin{aligned} Range & = Highest value - lowest value \\ Range & = 11 - 5 \\ Range & = 6 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Highest value - lowest value}\\ \text{Range} & = 11 - 5\\ \text{Range} & = 6 \end{aligned}$

Therefore, the range of the data set is 6.

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ID is: 2863 Seed is: 3399

The range

Consider the set of raw data:

\begin{matrix} {9; 11; 6; 11; 9; 6; 11; 7; 5} \end{matrix}

$\begin{array}{c}\{9; 11; 6; 11; 9; 6; 11; 7; 5\}\end{array}$

What is the range of the data set?

Answer: Range:

numeric

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

\begin{matrix} {5; 6; 6; 7; 9; 9; 11; 11; 11} \end{matrix}

$\begin{array}{c}\{5; 6; 6; 7; 9; 9; 11; 11; 11\}\end{array}$

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

Organise the data
Identify the lowest value
Identify the highest value and subtract them

\begin{aligned} Range & = Highest value - lowest value \\ Range & = 11 - 5 \\ Range & = 6 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Highest value - lowest value}\\ \text{Range} & = 11 - 5\\ \text{Range} & = 6 \end{aligned}$

Therefore, the range of the data set is 6.

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ID is: 2863 Seed is: 8165

The range

Consider the set of raw data:

\begin{matrix} {7; 10; 10; 6; 10; 6; 11} \end{matrix}

$\begin{array}{c}\{7; 10; 10; 6; 10; 6; 11\}\end{array}$

What is the range of the data set?

Answer: Range:

numeric

2 attempts remaining

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

\begin{matrix} {6; 6; 7; 10; 10; 10; 11} \end{matrix}

$\begin{array}{c}\{6; 6; 7; 10; 10; 10; 11\}\end{array}$

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

Organise the data
Identify the lowest value
Identify the highest value and subtract them

\begin{aligned} Range & = Highest value - lowest value \\ Range & = 11 - 6 \\ Range & = 5 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Highest value - lowest value}\\ \text{Range} & = 11 - 6\\ \text{Range} & = 5 \end{aligned}$

Therefore, the range of the data set is 5.

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ID is: 855 Seed is: 8732

Calculating the range of a data set

A group of $\text{20}$ students count the number of stones they each have. These are the data they collect:

\begin{matrix} 1 12 7 5 15 \\ 6 14 11 11 12 \\ 5 4 16 13 5 \\ 18 1 5 6 7 \end{matrix}

$\begin{array}{c} 1 \; \; 12 \; \; 7 \; \; 5 \; \; 15 \; \; \\ 6 \; \; 14 \; \; 11 \; \; 11 \; \; 12 \; \; \\ 5 \; \; 4 \; \; 16 \; \; 13 \; \; 5 \; \; \\ 18 \; \; 1 \; \; 5 \; \; 6 \; \; 7 \; \; \end{array}$

Calculate the range of values in the data set.

Answer: The range of the data is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.

STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

\begin{matrix} Maximum value = 18 \end{matrix}

$\begin{array}{c}\text{Maximum value } = \text{18}\end{array}$

Then we find the minimum value in the data set.

\begin{matrix} Minimum value = 1 \end{matrix}

$\begin{array}{c}\text{Minimum value } = \text{1}\end{array}$

Finally, we calculate the range of the data set.

\begin{aligned} Range & = Maximum - Minimum \\ = (18) - (1) \\ = 17 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Maximum} - \text{Minimum}\\ & = (\text{18}) - (\text{1}) \\ & = \text{17} \end{aligned}$

So the range of the data is $\text{17}$ .

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ID is: 855 Seed is: 1037

Calculating the range of a data set

A group of $\text{10}$ students count the number of coins they each have. These are the data they collect:

\begin{matrix} 1 4 3 3 2 \\ 6 1 2 3 6 \end{matrix}

$\begin{array}{c} 1 \; \; 4 \; \; 3 \; \; 3 \; \; 2 \; \; \\ 6 \; \; 1 \; \; 2 \; \; 3 \; \; 6 \; \; \end{array}$

Calculate the range of values in the data set.

Answer: The range of the data is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.

STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

\begin{matrix} Maximum value = 6 \end{matrix}

$\begin{array}{c}\text{Maximum value } = \text{6}\end{array}$

Then we find the minimum value in the data set.

\begin{matrix} Minimum value = 1 \end{matrix}

$\begin{array}{c}\text{Minimum value } = \text{1}\end{array}$

Finally, we calculate the range of the data set.

\begin{aligned} Range & = Maximum - Minimum \\ = (6) - (1) \\ = 5 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Maximum} - \text{Minimum}\\ & = (\text{6}) - (\text{1}) \\ & = \text{5} \end{aligned}$

So the range of the data is $\text{5}$ .

Submit your answer as:

ID is: 855 Seed is: 1605

Calculating the range of a data set

A group of $\text{10}$ students count the number of sweets they each have. These are the data they collect:

\begin{matrix} 9 6 10 8 8 \\ 7 7 4 3 4 \end{matrix}

$\begin{array}{c} 9 \; \; 6 \; \; 10 \; \; 8 \; \; 8 \; \; \\ 7 \; \; 7 \; \; 4 \; \; 3 \; \; 4 \; \; \end{array}$

Calculate the range of values in the data set.

Answer: The range of the data is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.

STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

\begin{matrix} Maximum value = 10 \end{matrix}

$\begin{array}{c}\text{Maximum value } = \text{10}\end{array}$

Then we find the minimum value in the data set.

\begin{matrix} Minimum value = 3 \end{matrix}

$\begin{array}{c}\text{Minimum value } = \text{3}\end{array}$

Finally, we calculate the range of the data set.

\begin{aligned} Range & = Maximum - Minimum \\ = (10) - (3) \\ = 7 \end{aligned}

$\begin{aligned} \text{Range} & = \text{Maximum} - \text{Minimum}\\ & = (\text{10}) - (\text{3}) \\ & = \text{7} \end{aligned}$

So the range of the data is $\text{7}$ .

Submit your answer as:

3. Interquartile range & percentiles

4. Practical applications

ID is: 1294 Seed is: 524

Application of population standard deviation

Three students measure how many cars drive by their house every day for $\text{1}$ week ( $\text{7}$ days). This is the data they collect:

Student A:

M	Tu	W	Th	F	Sa	Su
$\text{10}$	$\text{12}$	$\text{13}$	$\text{5}$	$\text{14}$	$\text{14}$	$\text{13}$

Student B:

M	Tu	W	Th	F	Sa	Su
$\text{6}$	$\text{8}$	$\text{21}$	$\text{7}$	$\text{4}$	$\text{2}$	$\text{17}$

Student C:

M	Tu	W	Th	F	Sa	Su
$\text{19}$	$\text{6}$	$\text{17}$	$\text{12}$	$\text{1}$	$\text{7}$	$\text{21}$

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.

STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

The "most consistent readings" corresponds to the standard deviation: the most consistent values will have the smallest standard deviation because the values are less spread out than the other sets. So we want to find the data set with the smallest standard deviation.

NOTE: Most calculators can calculate the standard deviation of a data set for you. It is important to be able to user your calculator to do this, because it will save you time and reduces the chances of errors in the calculation. The solution below shows some of the details of each standard deviation calculation, but if you use your calculator that is fine.

To determine the standard deviation of a data set, we need the number of values in the set ( $N$ ), and we need the mean for the data. For all three of these data sets $N = 7$ .

Now calculate the mean, $\bar{x}$ , for each data set, staring with Student A's data:

\begin{aligned} {\bar{x}}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ = \frac{10 + 12 + 13 + 5 + 14 + 14 + 13}{7} \\ = 11.57142... \end{aligned}

$\begin{aligned} \bar{x}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ & = \frac{10+12+13+5+14+14+13}{7} \\ & = \text{11.57142...} \end{aligned}$

For Student B:

\begin{aligned} {\bar{x}}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ = \frac{6 + 8 + 21 + 7 + 4 + 2 + 17}{7} \\ = 9.28571... \end{aligned}

$\begin{aligned} \bar{x}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ & = \frac{6+8+21+7+4+2+17}{7} \\ & = \text{9.28571...} \end{aligned}$

And for Student C:

\begin{aligned} {\bar{x}}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ = \frac{19 + 6 + 17 + 12 + 1 + 7 + 21}{7} \\ = 11.85714... \end{aligned}

$\begin{aligned} \bar{x}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ & = \frac{19+6+17+12+1+7+21}{7} \\ & = \text{11.85714...} \end{aligned}$

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

\begin{aligned} σ_{A} & = \sqrt{\frac{\sum {(x_{A} - {\bar{x}}_{A})}^{2}}{N_{A}}} \\ = \sqrt{\frac{\sum (x_{A} - 11.57142...)^{2}}{7}} \\ = \sqrt{8.81632...} \\ = 2.96922... \\ \approx 2.97 \end{aligned}

$\begin{aligned} \sigma_{A} & = \sqrt{\frac{\sum \left( x_{A} - \bar{x}_{A} \right)^2 }{N_{A}}} \\ & = \sqrt{\frac{\sum ( x_{A} - \text{11.57142...} )^2 }{\text{7}}} \\ & = \sqrt{\text{8.81632...}} \\ & = \text{2.96922...} \\ & \approx \text{2.97} \end{aligned}$

For Student B:

\begin{aligned} σ_{B} & = \sqrt{\frac{\sum {(x_{B} - {\bar{x}}_{B})}^{2}}{N_{B}}} \\ = \sqrt{\frac{\sum (x_{B} - 9.28571...)^{2}}{7}} \\ = \sqrt{42.20408...} \\ = 6.49646... \\ \approx 6.50 \end{aligned}

$\begin{aligned} \sigma_{B} & = \sqrt{\frac{\sum \left( x_{B} - \bar{x}_{B} \right)^2 }{N_{B}}} \\ & = \sqrt{\frac{\sum ( x_{B} - \text{9.28571...} )^2 }{\text{7}}} \\ & = \sqrt{\text{42.20408...}} \\ & = \text{6.49646...} \\ & \approx \text{6.50} \end{aligned}$

And for Student C:

\begin{aligned} σ_{C} & = \sqrt{\frac{\sum {(x_{C} - {\bar{x}}_{C})}^{2}}{N_{C}}} \\ = \sqrt{\frac{\sum (x_{C} - 11.85714...)^{2}}{7}} \\ = \sqrt{48.12244...} \\ = 6.93703... \\ \approx 6.94 \end{aligned}

$\begin{aligned} \sigma_{C} & = \sqrt{\frac{\sum \left( x_{C} - \bar{x}_{C} \right)^2 }{N_{C}}} \\ & = \sqrt{\frac{\sum ( x_{C} - \text{11.85714...} )^2 }{\text{7}}} \\ & = \sqrt{\text{48.12244...}} \\ & = \text{6.93703...} \\ & \approx \text{6.94} \end{aligned}$

Standard deviation tells us about the spread of a data set. A larger standard deviation value indicates data values that are more spread out (from each other). And a lower standard deviation indicates data values that are packed close to each other in a small range. So the smallest standard deviation for the three students indicates the most consistent data. Comparing the three results:

\begin{aligned} Student A: & σ_{A} \approx 2.97 \\ Student B: & σ_{B} \approx 6.50 \\ Student C: & σ_{C} \approx 6.94 \end{aligned}

$\begin{aligned} \text{Student A:} & \quad \sigma_{A} \approx \text{2.97} \\ \text{Student B:} & \quad \sigma_{B} \approx \text{6.50} \\ \text{Student C:} & \quad \sigma_{C} \approx \text{6.94} \end{aligned}$

Therefore the data set with the most consistent values is A.

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ID is: 1294 Seed is: 4277

Application of population standard deviation

Three students measure how many cars drive by their house every day for $\text{1}$ week ( $\text{7}$ days). This is the data they collect:

Student A:

M	Tu	W	Th	F	Sa	Su
$\text{14}$	$\text{8}$	$\text{10}$	$\text{15}$	$\text{20}$	$\text{20}$	$\text{15}$

Student B:

M	Tu	W	Th	F	Sa	Su
$\text{23}$	$\text{20}$	$\text{15}$	$\text{13}$	$\text{4}$	$\text{19}$	$\text{15}$

Student C:

M	Tu	W	Th	F	Sa	Su
$\text{14}$	$\text{4}$	$\text{14}$	$\text{22}$	$\text{20}$	$\text{5}$	$\text{6}$

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.

STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

To determine the standard deviation of a data set, we need the number of values in the set ( $N$ ), and we need the mean for the data. For all three of these data sets $N = 7$ .

Now calculate the mean, $\bar{x}$ , for each data set, staring with Student A's data:

\begin{aligned} {\bar{x}}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ = \frac{14 + 8 + 10 + 15 + 20 + 20 + 15}{7} \\ = 14.57142... \end{aligned}

$\begin{aligned} \bar{x}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ & = \frac{14+8+10+15+20+20+15}{7} \\ & = \text{14.57142...} \end{aligned}$

For Student B:

\begin{aligned} {\bar{x}}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ = \frac{23 + 20 + 15 + 13 + 4 + 19 + 15}{7} \\ = 15.57142... \end{aligned}

$\begin{aligned} \bar{x}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ & = \frac{23+20+15+13+4+19+15}{7} \\ & = \text{15.57142...} \end{aligned}$

And for Student C:

\begin{aligned} {\bar{x}}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ = \frac{14 + 4 + 14 + 22 + 20 + 5 + 6}{7} \\ = 12.14285... \end{aligned}

$\begin{aligned} \bar{x}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ & = \frac{14+4+14+22+20+5+6}{7} \\ & = \text{12.14285...} \end{aligned}$

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

\begin{aligned} σ_{A} & = \sqrt{\frac{\sum {(x_{A} - {\bar{x}}_{A})}^{2}}{N_{A}}} \\ = \sqrt{\frac{\sum (x_{A} - 14.57142...)^{2}}{7}} \\ = \sqrt{17.67346...} \\ = 4.20398... \\ \approx 4.20 \end{aligned}

$\begin{aligned} \sigma_{A} & = \sqrt{\frac{\sum \left( x_{A} - \bar{x}_{A} \right)^2 }{N_{A}}} \\ & = \sqrt{\frac{\sum ( x_{A} - \text{14.57142...} )^2 }{\text{7}}} \\ & = \sqrt{\text{17.67346...}} \\ & = \text{4.20398...} \\ & \approx \text{4.20} \end{aligned}$

For Student B:

\begin{aligned} σ_{B} & = \sqrt{\frac{\sum {(x_{B} - {\bar{x}}_{B})}^{2}}{N_{B}}} \\ = \sqrt{\frac{\sum (x_{B} - 15.57142...)^{2}}{7}} \\ = \sqrt{32.53061...} \\ = 5.70356... \\ \approx 5.70 \end{aligned}

$\begin{aligned} \sigma_{B} & = \sqrt{\frac{\sum \left( x_{B} - \bar{x}_{B} \right)^2 }{N_{B}}} \\ & = \sqrt{\frac{\sum ( x_{B} - \text{15.57142...} )^2 }{\text{7}}} \\ & = \sqrt{\text{32.53061...}} \\ & = \text{5.70356...} \\ & \approx \text{5.70} \end{aligned}$

And for Student C:

\begin{aligned} σ_{C} & = \sqrt{\frac{\sum {(x_{C} - {\bar{x}}_{C})}^{2}}{N_{C}}} \\ = \sqrt{\frac{\sum (x_{C} - 12.14285...)^{2}}{7}} \\ = \sqrt{45.83673...} \\ = 6.77028... \\ \approx 6.77 \end{aligned}

$\begin{aligned} \sigma_{C} & = \sqrt{\frac{\sum \left( x_{C} - \bar{x}_{C} \right)^2 }{N_{C}}} \\ & = \sqrt{\frac{\sum ( x_{C} - \text{12.14285...} )^2 }{\text{7}}} \\ & = \sqrt{\text{45.83673...}} \\ & = \text{6.77028...} \\ & \approx \text{6.77} \end{aligned}$

\begin{aligned} Student A: & σ_{A} \approx 4.20 \\ Student B: & σ_{B} \approx 5.70 \\ Student C: & σ_{C} \approx 6.77 \end{aligned}

$\begin{aligned} \text{Student A:} & \quad \sigma_{A} \approx \text{4.20} \\ \text{Student B:} & \quad \sigma_{B} \approx \text{5.70} \\ \text{Student C:} & \quad \sigma_{C} \approx \text{6.77} \end{aligned}$

Therefore the data set with the most consistent values is A.

Submit your answer as:

ID is: 1294 Seed is: 2035

Application of population standard deviation

Three students measure how many boys play soccer in the carpark every day for $\text{1}$ week ( $\text{7}$ days). This is the data they collect:

Student A:

M	Tu	W	Th	F	Sa	Su
$\text{20}$	$\text{14}$	$\text{18}$	$\text{27}$	$\text{15}$	$\text{11}$	$\text{13}$

Student B:

M	Tu	W	Th	F	Sa	Su
$\text{27}$	$\text{8}$	$\text{22}$	$\text{23}$	$\text{21}$	$\text{15}$	$\text{26}$

Student C:

M	Tu	W	Th	F	Sa	Su
$\text{18}$	$\text{13}$	$\text{23}$	$\text{5}$	$\text{20}$	$\text{19}$	$\text{9}$

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.

STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

To determine the standard deviation of a data set, we need the number of values in the set ( $N$ ), and we need the mean for the data. For all three of these data sets $N = 7$ .

Now calculate the mean, $\bar{x}$ , for each data set, staring with Student A's data:

\begin{aligned} {\bar{x}}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ = \frac{20 + 14 + 18 + 27 + 15 + 11 + 13}{7} \\ = 16.85714... \end{aligned}

$\begin{aligned} \bar{x}_{A} & = \frac{\sum (x_{A})}{N_{A}} \\ & = \frac{20+14+18+27+15+11+13}{7} \\ & = \text{16.85714...} \end{aligned}$

For Student B:

\begin{aligned} {\bar{x}}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ = \frac{27 + 8 + 22 + 23 + 21 + 15 + 26}{7} \\ = 20.28571... \end{aligned}

$\begin{aligned} \bar{x}_{B} & = \frac{\sum (x_{B})}{N_{B}} \\ & = \frac{27+8+22+23+21+15+26}{7} \\ & = \text{20.28571...} \end{aligned}$

And for Student C:

\begin{aligned} {\bar{x}}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ = \frac{18 + 13 + 23 + 5 + 20 + 19 + 9}{7} \\ = 15.28571... \end{aligned}

$\begin{aligned} \bar{x}_{C} & = \frac{\sum (x_{C})}{N_{C}} \\ & = \frac{18+13+23+5+20+19+9}{7} \\ & = \text{15.28571...} \end{aligned}$

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

\begin{aligned} σ_{A} & = \sqrt{\frac{\sum {(x_{A} - {\bar{x}}_{A})}^{2}}{N_{A}}} \\ = \sqrt{\frac{\sum (x_{A} - 16.85714...)^{2}}{7}} \\ = \sqrt{24.97959...} \\ = 4.99795... \\ \approx 5.00 \end{aligned}

$\begin{aligned} \sigma_{A} & = \sqrt{\frac{\sum \left( x_{A} - \bar{x}_{A} \right)^2 }{N_{A}}} \\ & = \sqrt{\frac{\sum ( x_{A} - \text{16.85714...} )^2 }{\text{7}}} \\ & = \sqrt{\text{24.97959...}} \\ & = \text{4.99795...} \\ & \approx \text{5.00} \end{aligned}$

For Student B:

\begin{aligned} σ_{B} & = \sqrt{\frac{\sum {(x_{B} - {\bar{x}}_{B})}^{2}}{N_{B}}} \\ = \sqrt{\frac{\sum (x_{B} - 20.28571...)^{2}}{7}} \\ = \sqrt{38.20408...} \\ = 6.18094... \\ \approx 6.18 \end{aligned}

$\begin{aligned} \sigma_{B} & = \sqrt{\frac{\sum \left( x_{B} - \bar{x}_{B} \right)^2 }{N_{B}}} \\ & = \sqrt{\frac{\sum ( x_{B} - \text{20.28571...} )^2 }{\text{7}}} \\ & = \sqrt{\text{38.20408...}} \\ & = \text{6.18094...} \\ & \approx \text{6.18} \end{aligned}$

And for Student C:

\begin{aligned} σ_{C} & = \sqrt{\frac{\sum {(x_{C} - {\bar{x}}_{C})}^{2}}{N_{C}}} \\ = \sqrt{\frac{\sum (x_{C} - 15.28571...)^{2}}{7}} \\ = \sqrt{36.20408...} \\ = 6.01698... \\ \approx 6.02 \end{aligned}

$\begin{aligned} \sigma_{C} & = \sqrt{\frac{\sum \left( x_{C} - \bar{x}_{C} \right)^2 }{N_{C}}} \\ & = \sqrt{\frac{\sum ( x_{C} - \text{15.28571...} )^2 }{\text{7}}} \\ & = \sqrt{\text{36.20408...}} \\ & = \text{6.01698...} \\ & \approx \text{6.02} \end{aligned}$

\begin{aligned} Student A: & σ_{A} \approx 5.00 \\ Student B: & σ_{B} \approx 6.18 \\ Student C: & σ_{C} \approx 6.02 \end{aligned}

$\begin{aligned} \text{Student A:} & \quad \sigma_{A} \approx \text{5.00} \\ \text{Student B:} & \quad \sigma_{B} \approx \text{6.18} \\ \text{Student C:} & \quad \sigma_{C} \approx \text{6.02} \end{aligned}$

Therefore the data set with the most consistent values is A.

Submit your answer as:

ID is: 3797 Seed is: 2554

Standard deviation: real world issues

HIV and AIDS

HIV damages a person's immune system by destroying a special type of blood cell called a CD4 cell. HIV-negative people (people without HIV) typically have a CD4 count ranging anywhere from $\text{500}$ up to $\text{1,500}$ . But HIV decreases the number of CD4 cells. If the CD4 count drops to $\text{200}$ or below, the person's immune system is critically weak, and he or she is diagnosed with AIDS. (Note that the CD4 count cannot be negative.) Answer the questions below about these two sets:

Set 1: CD4 count of people who are HIV-negative
Set 2: CD4 count of people who have been diagnosed with AIDS

Which of Set 1 or Set 2 would you expect to have the smaller standard deviation, and why?

A	Set 1 will have a smaller $\sigma$ -value because CD4 counts for HIV-negative people cover a much smaller range than for people diagnosed with AIDS.
B	Set 2 will have a smaller $\sigma$ -value because CD4 counts for HIV-negative people cover a much smaller range than for people diagnosed with AIDS.
C	Set 2 will have a smaller $\sigma$ -value because CD4 counts for HIV-negative people cover a much larger range than for people diagnosed with AIDS.
D	There is no reason to expect $\sigma$ be larger or smaller for either set.

Answer: The correct choice is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Which of the sets should have the smaller spread?

STEP: Imagine what the spread of each set will be like
[−1 point ⇒ 0 / 1 points left]

The set with the smaller standard deviation is the one we expect to be less spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

People who are HIV-negative have CD4 counts ranging from $\text{500}$ up to $\text{1,500}$ . So their CD4 counts are spread over a range of $\text{1,000}$ . In contrast, for people diagnosed with AIDS the CD4 count must be between $\text{0}$ and $\text{200}$ . So the standard deviation of results for those with AIDS will be smaller than the HIV-negative population.

NOTE: In recent years, anti-retroviral medications (ARVs) have made a massive impact on halting the advance and spread of HIV. ARVs can lead to a stable and even increasing CD4 count. HIV-positive individuals who take ARVs significantly reduce their chances of progressing to AIDS.

The correct answer is Set 2 will have a smaller $\sigma$ -value because CD4 counts for HIV-negative people cover a much larger range than for people diagnosed with AIDS, which is choice C.

Submit your answer as:

The numbers below are a small data set showing CD4 counts for ten people taken at random. Note that the numbers are given in hundreds ( $\text{3.5}$ means $\text{350}$ and $\text{12.2}$ means $\text{1,220}$ ).

$\begin{array}{c} \text{8.3} \; ; \; \text{11.2} \; ; \; \text{16.4} \; ; \; \text{12.4} \; ; \; \\ \text{11.0} \; ; \; \text{7.4} \; ; \; \text{6.4} \; ; \; \text{8.9} \; ; \; \\ \text{2.5} \; ; \; \text{6.5} \end{array}$

$\begin{array}{c}\text{Hundreds of cells}\end{array}$

Calculate the standard deviation of these values.
INSTRUCTIONS:
- You may use a calculator.
- Round your answer to two decimal places, if necessary.
- Give your answer in hundreds of cells (eg 3,45 for $\text{345}$ cells).
Answer: $\sigma \approx$ (hundreds of cells)
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

$\begin{aligned} \sigma = \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \end{aligned}$

STEP: Enter the data into your calculator
[−1 point ⇒ 1 / 2 points left]

Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

If you are not sure how to use your calculator to calculate statisitcs, try these options:
- search for a video about calculating statistics on your calculator
- ask a classmate who might know
- check with your teacher
TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be $N = 10$ .

STEP: Ask your calculator for the standard deviation
[−1 point ⇒ 0 / 2 points left]

Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

$\begin{aligned} \sigma &= \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \\ &= \text{3.65212...} \\ & \approx \text{3.65}\text{ (hundereds of cells)} \end{aligned}$
TIP:
You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for $\sigma$ , if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:
- typing a number incorrectly
- leaving out a number
- typing a number more than once
One way to check if your answer is reasonable is by comparing the $\sigma$ -value to the mean. (You can also ask your calculator for the mean.) Make sure that the $\text{10}$ data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

$\begin{aligned} \bar{x} = \text{9.1} \end{aligned}$
Rounded to two decimal places, the standard deviation is $\sigma \approx$ $\text{3.65}$ (hundreds of cells) .

Submit your answer as:
For many symmetric data sets, about $\text{68.3}\%$ of data values will be within one $\sigma$ of the mean. About $\text{95.4}\%$ will be within two $\sigma$ of the mean, and $\text{99.7}\%$ will be within three $\sigma$ of the mean. This is captured in the figure below (which is not precisely to scale):

We typically consider any value more than three $\sigma$ from the mean to be highly unusual. The mean of the $\text{10}$ CD4 count values given in Question 2 is $\text{9.1}$ . Use the answer from Question 2 to determine the upper boundary of the $3\sigma$ region and determine if $\text{13.36}$ (meaning $\text{1,336}$ CD4 cells) is highly unusual or not according to these data.
INSTRUCTIONS:
- Use the rounded answer for $\sigma$ from Question 2.
- Round your answer to two decimal places if necessary.
Answer:
- The upper boundary is .
- According to these data, a CD4 count of $\text{13.36}$ highly unusual.
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This example shows the lower boundary of the three sigma region:

$\begin{aligned} \text{lower limit}_{3\sigma} & = \bar{x} - 3\sigma \\ & = \left(\text{9.1}\right) - 3 \left(\text{3.65}\right)\\ & = -\text{1.85} \end{aligned}$

STEP: Compute the upper boundary
[−1 point ⇒ 1 / 2 points left]

The $3\sigma$ region spans from $3\sigma$ below the mean to $3\sigma$ above the mean. We want the upper limit of that region:

$\begin{aligned} \text{upper limit}_{3\sigma} & = \bar{x} + 3\sigma \\ & = \left( \text{9.1} \right) + 3 \left(\text{3.65}\right)\\ & = \text{20.05} \end{aligned}$

The upper boundary of the three sigma region is $\text{20.05}$ .

STEP: Compare this boundary to $\text{13.36}$ to answer the second part
[−1 point ⇒ 0 / 2 points left]

Now we can determine if $\text{13.36}$ is a highly unusual data value. The question states that data are highly unusual if they are more than $3\sigma$ from the mean. So the answer depends on whether the number is inside or outside of the $3\sigma$ region. The figure below shows that it is within the region.

Based on the image, we can see that for the data set in Question 2, $\text{13.36}$ is not highly unusual.

Submit your answer as: and

ID is: 3797 Seed is: 5958

Standard deviation: real world issues

Life expectancy

Life expectancy is the average number of years people live in a given country. It is an indicator of a country's overall development because more developed countries generally have higher life expectancy. In South Africa in 2015, life expectancy for men was $\text{59}$ years and life expectancy for women was $\text{66}$ years (source: World Health Organisation). Suppose that violence causes many more male deaths in younger years than female deaths (which is not necessarily true). Answer the questions below about these two sets:

Set 1: Age at death of women in SA
Set 2: Age at death of men in SA

Which of Set 1 or Set 2 would you expect to have the larger standard deviation, and why?

A	Set 1 will have a larger $\sigma$ -value because violence causes more young females to die than males.
B	Set 2 will have a larger $\sigma$ -value because violence causes more young males to die than females.
C	Set 2 will have a larger $\sigma$ -value because violence causes more young females to die than males.
D	There is no reason to expect $\sigma$ be larger or smaller for either set.

Answer: The correct choice is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Which of the sets should have the larger spread?

STEP: Imagine what the spread of each set will be like
[−1 point ⇒ 0 / 1 points left]

The set with the larger standard deviation is the one we expect to be more spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

The question says, "Suppose that violence causes many more male deaths in younger years than female deaths." If that is true, there are significantly more deaths for young men than for young women. That means more females survive to older ages. So the spread of ages at which males die would be wider than for females. This means we can expect that the standard deviation of the ages for the males should be larger.

NOTE: Violence has long been a leading cause of death in South Africa. According to The Lancet (as cited on News24 and BusinessTech), violence was the fifth leading cause of deaths in South Africa in 2013. (Other studies, covering different time periods, rank violence as the seventh or eighth leading cause of deaths.)

The correct answer is Set 2 will have a larger $\sigma$ -value because violence causes more young males to die than females, which is choice B.

Submit your answer as:

The numbers below are a small data set showing the age of ten people at the time of their deaths (in years).

$\begin{array}{c} \text{83} \; ; \; \text{80} \; ; \; \text{81} \; ; \; \text{71} \; ; \; \\ \text{76} \; ; \; \text{63} \; ; \; \text{20} \; ; \; \text{86} \; ; \; \\ \text{62} \; ; \; \text{44} \end{array}$

Calculate the standard deviation of these values.
INSTRUCTIONS:
- You may use a calculator.
- Round your answer to two decimal places, if necessary.
Answer: $\sigma \approx$ years
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

$\begin{aligned} \sigma = \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \end{aligned}$

STEP: Enter the data into your calculator
[−1 point ⇒ 1 / 2 points left]

Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

If you are not sure how to use your calculator to calculate statisitcs, try these options:
- search for a video about calculating statistics on your calculator
- ask a classmate who might know
- check with your teacher
TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be $N = 10$ .

STEP: Ask your calculator for the standard deviation
[−1 point ⇒ 0 / 2 points left]

Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

$\begin{aligned} \sigma &= \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \\ &= \text{19.63771...} \\ & \approx \text{19.64}\text{ years} \end{aligned}$
TIP:
You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for $\sigma$ , if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:
- typing a number incorrectly
- leaving out a number
- typing a number more than once
One way to check if your answer is reasonable is by comparing the $\sigma$ -value to the mean. (You can also ask your calculator for the mean.) Make sure that the $\text{10}$ data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

$\begin{aligned} \bar{x} = \text{66.6} \end{aligned}$
Rounded to two decimal places, the standard deviation is $\sigma \approx$ $\text{19.64}$ years .

Submit your answer as:
For many symmetric data sets, about $\text{68.3}\%$ of data values will be within one $\sigma$ of the mean. About $\text{95.4}\%$ will be within two $\sigma$ of the mean, and $\text{99.7}\%$ will be within three $\sigma$ of the mean. This is captured in the figure below (which is not precisely to scale):

We typically consider any value more than three $\sigma$ from the mean to be highly unusual. The mean of the $\text{10}$ ages given in Question 2 is $\text{66.6}$ . Use the answer from Question 2 to determine the lower boundary of the $3\sigma$ region and determine if dying at the age of $\text{47}$ years is highly unusual or not according to these data.
INSTRUCTIONS:
- Use the rounded answer for $\sigma$ from Question 2.
- Round your answer to two decimal places if necessary.
Answer:
- The lower boundary is .
- According to these data, death at the age of $\text{47}$ highly unusual.
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This example shows the upper boundary of the three sigma region:

$\begin{aligned} \text{upper limit}_{3\sigma} & = \bar{x} + 3\sigma \\ & = \left(\text{66.6}\right) + 3 \left(\text{19.64}\right)\\ & = \text{125.52} \end{aligned}$

STEP: Compute the lower boundary
[−1 point ⇒ 1 / 2 points left]

The $3\sigma$ region spans from $3\sigma$ below the mean to $3\sigma$ above the mean. We want the lower limit of that region:

$\begin{aligned} \text{lower limit}_{3\sigma} & = \bar{x} - 3\sigma \\ & = \left( \text{66.6} \right) - 3 \left(\text{19.64}\right)\\ & = \text{7.68} \end{aligned}$

The lower boundary of the three sigma region is $\text{7.68}$ .

STEP: Compare this boundary to $\text{47}$ to answer the second part
[−1 point ⇒ 0 / 2 points left]

Now we can determine if $\text{47}$ is a highly unusual data value. The question states that data are highly unusual if they are more than $3\sigma$ from the mean. So the answer depends on whether the number is inside or outside of the $3\sigma$ region. The figure below shows that it is within the region.

Based on the image, we can see that for the data set in Question 2, $\text{47}$ is not highly unusual.

Submit your answer as: and

ID is: 3797 Seed is: 2915

Standard deviation: real world issues

Income inequality

Income inequality is about how equally income is shared by people in a society. the Netherlands has a very low degree of income inequality, because people in the Netherlands earn similar amounts. In contrast, South Africa has one of the highest income inequalities in the world (source: The Organisation for Economic Co-operation and Development). Answer the questions below about these two sets:

Set 1: Incomes in South Africa
Set 2: Incomes in the Netherlands

Which of Set 1 or Set 2 would you expect to have the larger standard deviation, and why?

A	Set 1 will have a larger $\sigma$ -value because the gap between rich and poor in South Africa is smaller than in the Netherlands.
B	Set 1 will have a larger $\sigma$ -value because the gap between rich and poor in South Africa is larger than in the Netherlands.
C	Set 2 will have a larger $\sigma$ -value because the gap between rich and poor in South Africa is smaller than in the Netherlands.
D	There is no reason to expect $\sigma$ be larger or smaller for either set.

Answer: The correct choice is option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Which of the sets should have the larger spread?

STEP: Imagine what the spread of each set will be like
[−1 point ⇒ 0 / 1 points left]

The set with the larger standard deviation is the one we expect to be more spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

The question states that South Africa has higher income inequality than the Netherlands. So incomes for the rich and the poor in the Netherlands are not very different to each other. That means a small standard deviation. But in South Africa, incomes for rich and poor are significantly different. That means the gap between the rich and the poor in South Africa is larger than the gap in the Netherlands. This large gap would lead to a large standard deviation.

NOTE: Economists believe that education is a key element to reducing income inequality.

The correct answer is Set 1 will have a larger $\sigma$ -value because the gap between rich and poor in South Africa is larger than in the Netherlands, which is choice B.

Submit your answer as:

The numbers below are a small data set showing incomes for ten people in South Africa. Note that the numbers are given in thousands of rands (so $\text{4.2}$ means $\text{N}\llap{=}\,\text{4,200}$ ).

$\begin{array}{c} \text{5.9} \; ; \; \text{8.6} \; ; \; \text{7.1} \; ; \; \text{5.5} \; ; \; \\ \text{6.4} \; ; \; \text{33.7} \; ; \; \text{9.1} \; ; \; \text{6.6} \; ; \; \\ \text{17.2} \; ; \; \text{3.5} \end{array}$

$\begin{array}{c}\text{Thousands of rands}\end{array}$

Calculate the standard deviation of these values.
INSTRUCTIONS:
- You may use a calculator.
- Round your answer to two decimal places, if necessary.
- Give your answer in thousands of rands (eg, 4,32 for $\text{N}\llap{=}\,\text{4,320}$ ).
Answer: $\sigma \approx$ $\text{N}\llap{=}$ (in thousands)
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

$\begin{aligned} \sigma = \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \end{aligned}$

STEP: Enter the data into your calculator
[−1 point ⇒ 1 / 2 points left]

Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

If you are not sure how to use your calculator to calculate statisitcs, try these options:
- search for a video about calculating statistics on your calculator
- ask a classmate who might know
- check with your teacher
TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be $N = 10$ .

STEP: Ask your calculator for the standard deviation
[−1 point ⇒ 0 / 2 points left]

Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

$\begin{aligned} \sigma &= \sqrt{ \frac{\Sigma \left( x_i - \bar{x} \right)^2}{N} } \\ &= \text{8.52786...} \\ & \approx \text{N}\llap{=}\text{8.53}\text{ (in thousands)} \end{aligned}$
TIP:
You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for $\sigma$ , if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:
- typing a number incorrectly
- leaving out a number
- typing a number more than once
One way to check if your answer is reasonable is by comparing the $\sigma$ -value to the mean. (You can also ask your calculator for the mean.) Make sure that the $\text{10}$ data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

$\begin{aligned} \bar{x} = \text{10.36} \end{aligned}$
Rounded to two decimal places, the standard deviation is $\sigma \approx$ $\text{N}\llap{=}$ $\text{8.53}$ (in thousands) .

Submit your answer as:
For many symmetric data sets, about $\text{68.3}\%$ of data values will be within one $\sigma$ of the mean. About $\text{95.4}\%$ will be within two $\sigma$ of the mean, and $\text{99.7}\%$ will be within three $\sigma$ of the mean. This is captured in the figure below (which is not precisely to scale):

We typically consider any value more than three $\sigma$ from the mean to be highly unusual. The mean of the $\text{10}$ incomes given in Question 2 is $\text{10.36}$ . Use the answer from Question 2 to determine the upper boundary of the $3\sigma$ region and determine if $\text{34.7}$ (meaning an income of $\text{N}\llap{=}\,\text{34,700}$ ) is highly unusual or not according to these data.
INSTRUCTIONS:
- Use the rounded answer for $\sigma$ from Question 2.
- Round your answer to two decimal places if necessary.
Answer:
- The upper boundary is .
- According to these data, an income of $\text{34.7}$ highly unusual.
one-of
type(numeric.abserror(0.005))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This example shows the lower boundary of the three sigma region:

$\begin{aligned} \text{lower limit}_{3\sigma} & = \bar{x} - 3\sigma \\ & = \left(\text{10.36}\right) - 3 \left(\text{8.53}\right)\\ & = -\text{15.23} \end{aligned}$

STEP: Compute the upper boundary
[−1 point ⇒ 1 / 2 points left]

The $3\sigma$ region spans from $3\sigma$ below the mean to $3\sigma$ above the mean. We want the upper limit of that region:

$\begin{aligned} \text{upper limit}_{3\sigma} & = \bar{x} + 3\sigma \\ & = \left( \text{10.36} \right) + 3 \left(\text{8.53}\right)\\ & = \text{35.95} \end{aligned}$

The upper boundary of the three sigma region is $\text{35.95}$ .

STEP: Compare this boundary to $\text{34.7}$ to answer the second part
[−1 point ⇒ 0 / 2 points left]

Now we can determine if $\text{34.7}$ is a highly unusual data value. The question states that data are highly unusual if they are more than $3\sigma$ from the mean. So the answer depends on whether the number is inside or outside of the $3\sigma$ region. The figure below shows that it is within the region.

Based on the image, we can see that for the data set in Question 2, $\text{34.7}$ is not highly unusual.

Submit your answer as: and

ID is: 3784 Seed is: 5970

Real world statistics

The data below show the number of people at a movie theatre each day to watch a particular movie.

\begin{matrix} 26; 32; 36; 39; 47; \\ 26; 39; 31; 34; 35; \\ 45; 30; 50; 34; 26; \\ 34; 34; 30; 32; 42 \end{matrix}

$\begin{array}{c} 26; \; \;32; \; \;36; \; \;39; \; \;47; \; \; \\ 26; \; \;39; \; \;31; \; \;34; \; \;35; \; \; \\ 45; \; \;30; \; \;50; \; \;34; \; \;26; \; \; \\ 34; \; \;34; \; \;30; \; \;32; \; \;42 \end{array}$

Determine the mean of the data.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: $\bar{x} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The formula for the mean is:

$\begin{array}{c} \bar{x} = \dfrac{\text{sum of values}}{N} \end{array}$

STEP: Compute the mean
[−2 points ⇒ 0 / 2 points left]

The mean of the data is:

$\begin{aligned} \bar{x} & = \dfrac{\text{sum of values}}{N} \\ & = \dfrac{\text{702}}{\text{20}} \\ & = \text{35.1} \end{aligned}$

TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is $\text{50}$ , the mean should be smaller than $\text{50}$ .

The mean of the data is $\bar{x} = \text{35.1}$ .

Submit your answer as:
Calculate the standard deviation of the data.

INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.

Answer: $\sigma =$
one-of
type(numeric.abserror(0.005))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the formula for standard deviation. Then substitute in the values.

STEP: Substitute values into the formula
[−1 point ⇒ 1 / 2 points left]

The formula for the standard deviation is:

$\begin{array}{c} \sigma = \sqrt{\dfrac{\Sigma \left(x - \bar{x}\right)^{2}}{N}} \end{array}$

TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

We already know the values for $\bar{x}$ and $N$ . We also need to substitute in each of the data values:

$\begin{array}{c} 26; \; \;32; \; \;36; \; \;39; \; \;47; \; \; \\ 26; \; \;39; \; \;31; \; \;34; \; \;35; \; \; \\ 45; \; \;30; \; \;50; \; \;34; \; \;26; \; \; \\ 34; \; \;34; \; \;30; \; \;32; \; \;42 \end{array}$

Putting it all together:

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(26 - \text{35.1} \right)^{2} \\ + \left(32 - \text{35.1} \right)^{2} \\ + \ldots + \left(42 - \text{35.1}\right)^{2} \end{array}\right)}{\text{20}}} \end{aligned}$

STEP: Evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now we can work out the final answer.

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(26 - \text{35.1} \right)^{2} \\ + \left(32 - \text{35.1} \right)^{2} \\ + \ldots + \left(42 - \text{35.1}\right)^{2} \end{array}\right)}{\text{20}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(-\text{9.1}\right)^{2} + \left(-\text{3.1}\right)^{2} + \\ \ldots + \left(\text{6.9}\right)^{2}\end{array}}{\text{20}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(\text{82.81}\right) + \left(\text{9.61}\right) + \\ \ldots + \left(\text{47.61}\right)\end{array}}{\text{20}}} \\ & = \sqrt{\frac{\text{881.8}}{\text{20}}} \\ & = \text{6.64003...}\\ & \approx \text{6.64} \end{aligned}$

TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

Rounded to two decimal places, the standard deviation is $\sigma = \text{6.64}$ .

Submit your answer as:

Determine the number of days for which the number of people watching the movie was within one standard deviation of the mean.

Answer:

The number of days is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the values one standard deviation above and below the mean.

STEP: Determine the upper and lower boundaries for one standard deviation from the mean
[−1 point ⇒ 1 / 2 points left]

To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

\begin{aligned} lower boundary & = 35.1 - 6.64 \\ = 28.46 \\ upper boundary & = 35.1 + 6.64 \\ = 41.74 \end{aligned}

$\begin{aligned} \text{lower boundary} & = \text{35.1} - \text{6.64} \\ & = \text{28.46} \\[0.2cm] \text{upper boundary} & = \text{35.1} + \text{6.64} \\ & = \text{41.74} \end{aligned}$

STEP: Count the values within

\bar{x} \pm σ

$\bar{x} \pm \sigma$
[−1 point ⇒ 0 / 2 points left]

Now we must count how many of the values there are from $\text{28.46}$ up to $\text{41.74}$ . It is helpful to organise the data into a table to separate the values we want from the ones we don't.

$n < \text{28.46}$	$\begin{array}{l}26; \;26; \;26 \;\end{array}$
$\text{28.46} \;$ $\leq n \leq$ $\; \text{41.74}$	$\begin{array}{l}32; \;36; \;39; \;39; \; \\ 31; \;34; \;35; \;30; \; \\ 34; \;34; \;34; \;30; \; \\ 32 \;\end{array}$
$n > \text{41.74}$	$\begin{array}{l}47; \;45; \;50; \;42 \; \\ \end{array}$

The number of days for which the number of people watching the movie was within one standard deviation of the mean value is $\text{13}$ .

Submit your answer as:

ID is: 3784 Seed is: 7949

Real world statistics

The data below show the number of people visiting a farm per day to ride horses.

\begin{matrix} 13; 3; 2; 8; 7; \\ 4; 2; 6; 9; 6; \\ 12; 12; 15; 10; 8; \\ 7; 13; 9 \end{matrix}

$\begin{array}{c} 13; \; \;3; \; \;2; \; \;8; \; \;7; \; \; \\ 4; \; \;2; \; \;6; \; \;9; \; \;6; \; \; \\ 12; \; \;12; \; \;15; \; \;10; \; \;8; \; \; \\ 7; \; \;13; \; \;9 \end{array}$

Determine the mean of the data.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: $\bar{x} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The formula for the mean is:

$\begin{array}{c} \bar{x} = \dfrac{\text{sum of values}}{N} \end{array}$

STEP: Compute the mean
[−2 points ⇒ 0 / 2 points left]

The mean of the data is:

$\begin{aligned} \bar{x} & = \dfrac{\text{sum of values}}{N} \\ & = \dfrac{\text{146}}{\text{18}} \\ & = \text{8.11111...} \\ & \approx \text{8.11} \end{aligned}$

TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is $\text{15}$ , the mean should be smaller than $\text{15}$ .

The mean of the data is $\bar{x} = \text{8.11}$ .

Submit your answer as:
Calculate the standard deviation of the data.

INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.

Answer: $\sigma =$
one-of
type(numeric.abserror(0.005))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the formula for standard deviation. Then substitute in the values.

STEP: Substitute values into the formula
[−1 point ⇒ 1 / 2 points left]

The formula for the standard deviation is:

$\begin{array}{c} \sigma = \sqrt{\dfrac{\Sigma \left(x - \bar{x}\right)^{2}}{N}} \end{array}$

TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

We already know the values for $\bar{x}$ and $N$ . We also need to substitute in each of the data values:

$\begin{array}{c} 13; \; \;3; \; \;2; \; \;8; \; \;7; \; \; \\ 4; \; \;2; \; \;6; \; \;9; \; \;6; \; \; \\ 12; \; \;12; \; \;15; \; \;10; \; \;8; \; \; \\ 7; \; \;13; \; \;9 \end{array}$

Putting it all together:

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(13 - \text{8.11} \right)^{2} \\ + \left(3 - \text{8.11} \right)^{2} \\ + \ldots + \left(9 - \text{8.11}\right)^{2} \end{array}\right)}{\text{18}}} \end{aligned}$

STEP: Evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now we can work out the final answer.

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(13 - \text{8.11} \right)^{2} \\ + \left(3 - \text{8.11} \right)^{2} \\ + \ldots + \left(9 - \text{8.11}\right)^{2} \end{array}\right)}{\text{18}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(\text{4.89}\right)^{2} + \left(-\text{5.11}\right)^{2} + \\ \ldots + \left(\text{0.89}\right)^{2}\end{array}}{\text{18}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(\text{23.9121}\right) + \left(\text{26.1121}\right) + \\ \ldots + \left(\text{0.7921}\right)\end{array}}{\text{18}}} \\ & = \sqrt{\frac{\text{259.7778}}{\text{18}}} \\ & = \text{3.79896...}\\ & \approx \text{3.8} \end{aligned}$

TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

Rounded to two decimal places, the standard deviation is $\sigma = \text{3.8}$ .

Submit your answer as:

Determine the number of days for which the number of people riding horses was within one standard deviation of the mean.

Answer:

The number of days is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the values one standard deviation above and below the mean.

STEP: Determine the upper and lower boundaries for one standard deviation from the mean
[−1 point ⇒ 1 / 2 points left]

To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

\begin{aligned} lower boundary & = 8.11 - 3.8 \\ = 4.31 \\ upper boundary & = 8.11 + 3.8 \\ = 11.91 \end{aligned}

$\begin{aligned} \text{lower boundary} & = \text{8.11} - \text{3.8} \\ & = \text{4.31} \\[0.2cm] \text{upper boundary} & = \text{8.11} + \text{3.8} \\ & = \text{11.91} \end{aligned}$

STEP: Count the values within

\bar{x} \pm σ

$\bar{x} \pm \sigma$
[−1 point ⇒ 0 / 2 points left]

Now we must count how many of the values there are from $\text{4.31}$ up to $\text{11.91}$ . It is helpful to organise the data into a table to separate the values we want from the ones we don't.

$n < \text{4.31}$	$\begin{array}{l}3; \;2; \;4; \;2 \; \\ \end{array}$
$\text{4.31} \;$ $\leq n \leq$ $\; \text{11.91}$	$\begin{array}{l}8; \;7; \;6; \;9; \; \\ 6; \;10; \;8; \;7; \; \\ 9 \;\end{array}$
$n > \text{11.91}$	$\begin{array}{l}13; \;12; \;12; \;15; \; \\ 13 \;\end{array}$

The number of days for which the number of people riding horses was within one standard deviation of the mean value is $\text{9}$ .

Submit your answer as:

ID is: 3784 Seed is: 7694

Real world statistics

The data below show the number of people going to Muizenberg per day to go surfing.

\begin{matrix} 69; 65; 64; 54; 64; \\ 62; 59; 53; 65; 41; \\ 64; 63; 50; 58; 70; \\ 59; 49; 59 \end{matrix}

$\begin{array}{c} 69; \; \;65; \; \;64; \; \;54; \; \;64; \; \; \\ 62; \; \;59; \; \;53; \; \;65; \; \;41; \; \; \\ 64; \; \;63; \; \;50; \; \;58; \; \;70; \; \; \\ 59; \; \;49; \; \;59 \end{array}$

Determine the mean of the data.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: $\bar{x} =$
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The formula for the mean is:

$\begin{array}{c} \bar{x} = \dfrac{\text{sum of values}}{N} \end{array}$

STEP: Compute the mean
[−2 points ⇒ 0 / 2 points left]

The mean of the data is:

$\begin{aligned} \bar{x} & = \dfrac{\text{sum of values}}{N} \\ & = \dfrac{\text{1,068}}{\text{18}} \\ & = \text{59.33333...} \\ & \approx \text{59.33} \end{aligned}$

TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is $\text{70}$ , the mean should be smaller than $\text{70}$ .

The mean of the data is $\bar{x} = \text{59.33}$ .

Submit your answer as:
Calculate the standard deviation of the data.

INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.

Answer: $\sigma =$
one-of
type(numeric.abserror(0.005))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the formula for standard deviation. Then substitute in the values.

STEP: Substitute values into the formula
[−1 point ⇒ 1 / 2 points left]

The formula for the standard deviation is:

$\begin{array}{c} \sigma = \sqrt{\dfrac{\Sigma \left(x - \bar{x}\right)^{2}}{N}} \end{array}$

TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

We already know the values for $\bar{x}$ and $N$ . We also need to substitute in each of the data values:

$\begin{array}{c} 69; \; \;65; \; \;64; \; \;54; \; \;64; \; \; \\ 62; \; \;59; \; \;53; \; \;65; \; \;41; \; \; \\ 64; \; \;63; \; \;50; \; \;58; \; \;70; \; \; \\ 59; \; \;49; \; \;59 \end{array}$

Putting it all together:

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(69 - \text{59.33} \right)^{2} \\ + \left(65 - \text{59.33} \right)^{2} \\ + \ldots + \left(59 - \text{59.33}\right)^{2} \end{array}\right)}{\text{18}}} \end{aligned}$

STEP: Evaluate the answer
[−1 point ⇒ 0 / 2 points left]

Now we can work out the final answer.

$\begin{aligned} \sigma & = \sqrt{\frac{\left(\begin{array}{c}\left(69 - \text{59.33} \right)^{2} \\ + \left(65 - \text{59.33} \right)^{2} \\ + \ldots + \left(59 - \text{59.33}\right)^{2} \end{array}\right)}{\text{18}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(\text{9.67}\right)^{2} + \left(\text{5.67}\right)^{2} + \\ \ldots + \left(-\text{0.33}\right)^{2}\end{array}}{\text{18}}} \\ & = \sqrt{\frac{\begin{array}{c}\left(\text{93.5089}\right) + \left(\text{32.1489}\right) + \\ \ldots + \left(\text{0.1089}\right)\end{array}}{\text{18}}} \\ & = \sqrt{\frac{\text{958.0002}}{\text{18}}} \\ & = \text{7.29535...}\\ & \approx \text{7.3} \end{aligned}$

TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

Rounded to two decimal places, the standard deviation is $\sigma = \text{7.3}$ .

Submit your answer as:

Determine the number of days for which the number of people surfing at Muizenberg was within one standard deviation of the mean.

Answer:

The number of days is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the values one standard deviation above and below the mean.

STEP: Determine the upper and lower boundaries for one standard deviation from the mean
[−1 point ⇒ 1 / 2 points left]

To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

\begin{aligned} lower boundary & = 59.33 - 7.3 \\ = 52.03 \\ upper boundary & = 59.33 + 7.3 \\ = 66.63 \end{aligned}

$\begin{aligned} \text{lower boundary} & = \text{59.33} - \text{7.3} \\ & = \text{52.03} \\[0.2cm] \text{upper boundary} & = \text{59.33} + \text{7.3} \\ & = \text{66.63} \end{aligned}$

STEP: Count the values within

\bar{x} \pm σ

$\bar{x} \pm \sigma$
[−1 point ⇒ 0 / 2 points left]

Now we must count how many of the values there are from $\text{52.03}$ up to $\text{66.63}$ . It is helpful to organise the data into a table to separate the values we want from the ones we don't.

$n < \text{52.03}$	$\begin{array}{l}41; \;50; \;49 \;\end{array}$
$\text{52.03} \;$ $\leq n \leq$ $\; \text{66.63}$	$\begin{array}{l}65; \;64; \;54; \;64; \; \\ 62; \;59; \;53; \;65; \; \\ 64; \;63; \;58; \;59; \; \\ 59 \;\end{array}$
$n > \text{66.63}$	$\begin{array}{l}69; \;70 \;\end{array}$

The number of days for which the number of people surfing at Muizenberg was within one standard deviation of the mean value is $\text{13}$ .

Submit your answer as:

ID is: 4337 Seed is: 3028

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the weights (to the nearest kilogram) of each of the $\text{27}$ members of a medical trial.

\begin{array}{ccccc} 82 & 159 & 93 & 84 & 132 \\ 103 & 90 & 87 & 76 & 88 \\ 128 & 125 & 106 & 63 & 146 \\ 77 & 57 & 90 & 135 & 157 \\ 148 & 79 & 86 & 95 & 149 \\ 141 & 133 \end{array}

$\begin{array}{ccccc} 82&159&93&84&132\\103&90&87&76&88\\128&125&106&63&146\\77&57&90&135&157\\148&79&86&95&149\\141&133& \end{array}$

Write down the mode of the data.
Calculate the range of the data.
Determine the standard deviation of the data.

INSTRUCTION: Where appropriate, round your answer to two decimal places.

Answer:

The mode $=$ $\text{kg}$
The range $=$ $\text{kg}$
The standard deviation $\approx$ $\text{kg}$

numeric

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The mode is the data value that occurs most.
The range is the difference between the maximum and minimum data value.
The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

$\text{Standard deviation}$ $= \sqrt{\dfrac{\Sigma \left( x_i -\bar{x} \right)^2}{n}}$

STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]

The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value $\text{90}$ occurs twice, which is more times than any other data value.

Therefore the mode is $\text{90}\ \text{kg}$ .

STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]

The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

the maximum value is: $\text{159}\ \text{kg}$
the minimum value is: $\text{57}\ \text{kg}$

\begin{aligned} range = 159 - 57 & = 102 kg \end{aligned}

$\begin{aligned} \text{range} = \text{159} - \text{57} &= \text{102}\ \text{kg} \end{aligned}$

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

Put your calculator into STATS MODE
Type the data values into your calculator, and make sure you enter all $\text{27}$ values!
Use the $\sigma x$ function and then press $=$

Below is an example of the buttons to use on two particular calculators. Calculator functions change depending on the make and model you are using. You will need to look up the steps needed for your particular make and model of calculator.

SHARP calculator	CASIO calculator
STATS mode
1. press MODE	1. press MODE
2. press 1	2. press 2
3. press 0	3. press 1
Entering data values
value1, M+, value2, M+, ...	value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, =	SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

\begin{aligned} σ & = 30.14359... \\ \approx 30.14 kg \end{aligned}

$\begin{aligned} \sigma &= \text{30.14359...} \\ & \approx \text{30.14}\ \text{kg} \end{aligned}$

NOTE:

It is good to check that you have put in the data values correctly. It is easy to skip a number or type one of the values in twice. Get to know your calculator so that you can double-check that it has the correct number of data values once you have typed them in. Even though the calculator will not make a calculation error for $\sigma$ , if you made any mistakes typing the data into the calculator, the answer will be wrong.

STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean $(\overline{x})$ using the formula:

\begin{aligned} \bar{x} & = \frac{sum of values}{number of values in data set} \\ = \frac{2,909}{27} \\ = 107.74074... \\ \approx 107.74 kg \end{aligned}

$\begin{aligned} \overline{x} &= \dfrac{\text{sum of values}}{\text{number of values in data set}} \\ &= \dfrac{\text{2,909}}{\text{27}} \\ &= \text{107.74074...} \\ & \approx \text{107.74}\ \text{kg} \end{aligned}$

Fill out a table like this:

$\textbf{Data}$ $\textbf{values}$ $x$	$\textbf{(Deviations)}^2$ $(x-\overline{x})^2$
$\text{82}$	$(\text{82} - \text{107.74})^2$ $= \text{662.55}$
$\text{159}$	$(\text{159} - \text{107.74})^2$ $=\text{2,627.59}$
$\text{93}$	$(\text{93} - \text{107.74})^2$ $= \text{217.27}$
$\text{84}$	$(\text{84} - \text{107.74})^2$ $= \text{563.59}$
$\text{132}$	$(\text{132} - \text{107.74})^2$ $= \text{588.55}$
$...$	$...$
$\Sigma x$ $= \text{2,909}$	$\Sigma (x-\overline{x})^2$ $= \text{24,533.19}$

Now calculate the standard deviation $(\sigma)$ using the formula:

\begin{aligned} σ & = \sqrt{\frac{Σ {(x_{i} - \bar{x})}^{2}}{n}} \\ = 30.14359... \\ \approx 30.14 kg \end{aligned}

$\begin{aligned} \sigma &= \sqrt{\frac{\Sigma \left( x_i -\bar{x} \right)^2}{n}} \\ &= \text{30.14359...} \\ & \approx \text{30.14}\ \text{kg} \end{aligned}$

NOTE:

The statistics function on your calculator can store many exact values, even if the values have lots of numbers after the decimal comma (such as $\text{3.1415926...}$ ). The values shown in the table above are shown as rounded values. If you determine the standard deviation on paper using a table like this, your final answer may differ slightly because you will use rounded values instead of stored exact values.

Submit your answer as: andand

ID is: 4337 Seed is: 424

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the weights (to the nearest kilogram) of each of the $\text{29}$ members of a medical trial.

\begin{array}{ccccc} 114 & 156 & 93 & 157 & 137 \\ 74 & 96 & 128 & 98 & 78 \\ 120 & 118 & 152 & 71 & 87 \\ 75 & 81 & 104 & 62 & 105 \\ 127 & 96 & 131 & 111 & 129 \\ 141 & 139 & 89 & 143 \end{array}

$\begin{array}{ccccc} 114&156&93&157&137\\74&96&128&98&78\\120&118&152&71&87\\75&81&104&62&105\\127&96&131&111&129\\141&139&89&143& \end{array}$

Write down the mode of the data.
Calculate the range of the data.
Determine the standard deviation of the data.

INSTRUCTION: Where appropriate, round your answer to two decimal places.

Answer:

The mode $=$ $\text{kg}$
The range $=$ $\text{kg}$
The standard deviation $\approx$ $\text{kg}$

numeric

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The mode is the data value that occurs most.
The range is the difference between the maximum and minimum data value.
The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

$\text{Standard deviation}$ $= \sqrt{\dfrac{\Sigma \left( x_i -\bar{x} \right)^2}{n}}$

STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]

The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value $\text{96}$ occurs twice, which is more times than any other data value.

Therefore the mode is $\text{96}\ \text{kg}$ .

STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]

The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

the maximum value is: $\text{157}\ \text{kg}$
the minimum value is: $\text{62}\ \text{kg}$

\begin{aligned} range = 157 - 62 & = 95 kg \end{aligned}

$\begin{aligned} \text{range} = \text{157} - \text{62} &= \text{95}\ \text{kg} \end{aligned}$

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

Put your calculator into STATS MODE
Type the data values into your calculator, and make sure you enter all $\text{29}$ values!
Use the $\sigma x$ function and then press $=$

SHARP calculator	CASIO calculator
STATS mode
1. press MODE	1. press MODE
2. press 1	2. press 2
3. press 0	3. press 1
Entering data values
value1, M+, value2, M+, ...	value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, =	SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

\begin{aligned} σ & = 27.13650... \\ \approx 27.14 kg \end{aligned}

$\begin{aligned} \sigma &= \text{27.13650...} \\ & \approx \text{27.14}\ \text{kg} \end{aligned}$

NOTE:

STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean $(\overline{x})$ using the formula:

\begin{aligned} \bar{x} & = \frac{sum of values}{number of values in data set} \\ = \frac{3,212}{29} \\ = 110.75862... \\ \approx 110.76 kg \end{aligned}

$\begin{aligned} \overline{x} &= \dfrac{\text{sum of values}}{\text{number of values in data set}} \\ &= \dfrac{\text{3,212}}{\text{29}} \\ &= \text{110.75862...} \\ & \approx \text{110.76}\ \text{kg} \end{aligned}$

Fill out a table like this:

$\textbf{Data}$ $\textbf{values}$ $x$	$\textbf{(Deviations)}^2$ $(x-\overline{x})^2$
$\text{114}$	$(\text{114} - \text{110.76})^2$ $= \text{10.50}$
$\text{156}$	$(\text{156} - \text{110.76})^2$ $=\text{2,046.66}$
$\text{93}$	$(\text{93} - \text{110.76})^2$ $= \text{315.42}$
$\text{157}$	$(\text{157} - \text{110.76})^2$ $= \text{2,138.14}$
$\text{137}$	$(\text{137} - \text{110.76})^2$ $= \text{688.54}$
$...$	$...$
$\Sigma x$ $= \text{3,212}$	$\Sigma (x-\overline{x})^2$ $= \text{21,355.31}$

Now calculate the standard deviation $(\sigma)$ using the formula:

\begin{aligned} σ & = \sqrt{\frac{Σ {(x_{i} - \bar{x})}^{2}}{n}} \\ = 27.13650... \\ \approx 27.14 kg \end{aligned}

$\begin{aligned} \sigma &= \sqrt{\frac{\Sigma \left( x_i -\bar{x} \right)^2}{n}} \\ &= \text{27.13650...} \\ & \approx \text{27.14}\ \text{kg} \end{aligned}$

NOTE:

Submit your answer as: andand

ID is: 4337 Seed is: 6231

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of devices used by $\text{19}$ classes in a school for Siyavula Practice.

\begin{array}{ccccc} 25 & 10 & 6 & 15 & 23 \\ 5 & 14 & 19 & 24 & 4 \\ 16 & 5 & 7 & 13 & 3 \\ 17 & 1 & 8 & 2 \end{array}

$\begin{array}{ccccc} 25&10&6&15&23\\5&14&19&24&4\\16&5&7&13&3\\17&1&8&2& \end{array}$

Write down the mode of the data.
Calculate the range of the data.
Determine the standard deviation of the data.

INSTRUCTION: Where appropriate, round your answer to two decimal places.

Answer:

The mode $=$ $\text{devices}$
The range $=$ $\text{devices}$
The standard deviation $\approx$ $\text{devices}$

numeric

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The mode is the data value that occurs most.
The range is the difference between the maximum and minimum data value.
The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

$\text{Standard deviation}$ $= \sqrt{\dfrac{\Sigma \left( x_i -\bar{x} \right)^2}{n}}$

STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]

The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value $\text{5}$ occurs twice, which is more times than any other data value.

Therefore the mode is $\text{5}\ \text{devices}$ .

STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]

The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

the maximum value is: $\text{25}\ \text{devices}$
the minimum value is: $\text{1}\ \text{devices}$

\begin{aligned} range = 25 - 1 & = 24 devices \end{aligned}

$\begin{aligned} \text{range} = \text{25} - \text{1} &= \text{24}\ \text{devices} \end{aligned}$

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

Put your calculator into STATS MODE
Type the data values into your calculator, and make sure you enter all $\text{19}$ values!
Use the $\sigma x$ function and then press $=$

SHARP calculator	CASIO calculator
STATS mode
1. press MODE	1. press MODE
2. press 1	2. press 2
3. press 0	3. press 1
Entering data values
value1, M+, value2, M+, ...	value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, =	SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

\begin{aligned} σ & = 7.52760... \\ \approx 7.53 devices \end{aligned}

$\begin{aligned} \sigma &= \text{7.52760...} \\ & \approx \text{7.53}\ \text{devices} \end{aligned}$

NOTE:

STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean $(\overline{x})$ using the formula:

\begin{aligned} \bar{x} & = \frac{sum of values}{number of values in data set} \\ = \frac{217}{19} \\ = 11.42105... \\ \approx 11.42 devices \end{aligned}

$\begin{aligned} \overline{x} &= \dfrac{\text{sum of values}}{\text{number of values in data set}} \\ &= \dfrac{\text{217}}{\text{19}} \\ &= \text{11.42105...} \\ & \approx \text{11.42}\ \text{devices} \end{aligned}$

Fill out a table like this:

$\textbf{Data}$ $\textbf{values}$ $x$	$\textbf{(Deviations)}^2$ $(x-\overline{x})^2$
$\text{25}$	$(\text{25} - \text{11.42})^2$ $= \text{184.42}$
$\text{10}$	$(\text{10} - \text{11.42})^2$ $=\text{2.02}$
$\text{6}$	$(\text{6} - \text{11.42})^2$ $= \text{29.38}$
$\text{15}$	$(\text{15} - \text{11.42})^2$ $= \text{12.82}$
$\text{23}$	$(\text{23} - \text{11.42})^2$ $= \text{134.10}$
$...$	$...$
$\Sigma x$ $= \text{217}$	$\Sigma (x-\overline{x})^2$ $= \text{1,076.63}$

Now calculate the standard deviation $(\sigma)$ using the formula:

\begin{aligned} σ & = \sqrt{\frac{Σ {(x_{i} - \bar{x})}^{2}}{n}} \\ = 7.52760... \\ \approx 7.53 devices \end{aligned}

$\begin{aligned} \sigma &= \sqrt{\frac{\Sigma \left( x_i -\bar{x} \right)^2}{n}} \\ &= \text{7.52760...} \\ & \approx \text{7.53}\ \text{devices} \end{aligned}$

NOTE:

Submit your answer as: andand

A	Set 1 has a smaller mean but the standard deviation values are similar.
B	The mean values are similar but Set 1 has a smaller standard deviation.
C	The mean and standard deviation values for the sets are both similar.
D	Neither the mean nor the standard deviation values for the two sets are similar.

A	Neither the mean nor the standard deviation values for the two sets are similar.
B	Set 1 has a smaller mean but the standard deviation values are similar.
C	The mean and standard deviation values for the sets are both similar.
D	The mean values are similar but Set 1 has a smaller standard deviation.

A	The mean values are similar but Set 1 has a smaller standard deviation.
B	The mean and standard deviation values for the sets are both similar.
C	Set 1 has a larger mean but the standard deviation values are similar.
D	Neither the mean nor the standard deviation values for the two sets are similar.

A	$\begin{aligned} \bar{x} & \approx \text{120} \\ \sigma & \approx \text{25} \end{aligned}$
B	$\begin{aligned} \bar{x} & \approx \text{61} \\ \sigma & \approx \text{50} \end{aligned}$
C	$\begin{aligned} \bar{x} & \approx \text{30} \\ \sigma & \approx \text{50} \end{aligned}$
D	$\begin{aligned} \bar{x} & \approx \text{61} \\ \sigma & \approx \text{8} \end{aligned}$

A	$\begin{aligned} \bar{x} & \approx \text{93} \\ \sigma & \approx \text{37.5} \end{aligned}$
B	$\begin{aligned} \bar{x} & \approx \text{55} \\ \sigma & \approx \text{37.5} \end{aligned}$
C	$\begin{aligned} \bar{x} & \approx \text{93} \\ \sigma & \approx \text{13} \end{aligned}$
D	$\begin{aligned} \bar{x} & \approx \text{40} \\ \sigma & \approx \text{75} \end{aligned}$

A	The mean will be $\text{6} \ \text{cm}$ too small.
B	The mean will be $\text{94} \ \text{cm}$ too large.
C	The mean will be $\text{94} \ \text{cm}$ too small.
D	The error has no effect on the mean.

A	The standard deviation will be $\text{6} \ \text{cm}$ too small.
B	The standard deviation will be $\text{94} \ \text{cm}$ too large.
C	The error has no effect on the standard deviation.
D	The standard deviation will change by an unknown amount.

A	The mean will be $\text{0.1} \ \text{m}$ too large.
B	The mean will be $\text{0.1} \ \text{m}$ too small.
C	The mean will be $\text{0.9} \ \text{m}$ too large.
D	The error has no effect on the mean.

A	The standard deviation will be $\text{0.1} \ \text{m}$ too small.
B	The standard deviation will be $\text{0.9} \ \text{m}$ too large.
C	The error has no effect on the standard deviation.
D	The standard deviation will change by an unknown amount.

A	The mean will be $\text{106} \ \text{cm}$ too small.
B	The mean will be $\text{6} \ \text{cm}$ too large.
C	The mean will be $\text{6} \ \text{cm}$ too small.
D	The error has no effect on the mean.

A	The standard deviation will be $\text{106} \ \text{cm}$ too large.
B	The standard deviation will be $\text{106} \ \text{cm}$ too small.
C	The error has no effect on the standard deviation.
D	The standard deviation will change by an unknown amount.

A ✘	$\text{128}$
The spread of the data is much smaller than $\text{128}$ .
B ✘	$\text{3}$
The answer should be somewhat smaller than $\text{3}$ because the standard deviation does not span the entire data set.
C ✘	$\text{0}$
The answer cannot be zero because the data are not all equal to each other.
D ✔	$\text{1.18}$
This is the best match for the figure above.

A ✔	$\text{2.9}$
This is the best match for the figure above.
B ✘	$\text{121}$
The spread of the data is much smaller than $\text{121}$ .
C ✘	$\text{9}$
The answer should be somewhat smaller than $\text{9}$ because the standard deviation does not span the entire data set.
D ✘	$\text{126.8}$
The spread of the data is much smaller than $\text{126.8}$ .

A ✘	$\text{135.17}$
The spread of the data is much smaller than $\text{135.17}$ .
B ✘	$\text{1.34}$
This number is too small, because it implies the values are packed very close to the mean.
C ✔	$\text{7.03}$
This is the best match for the figure above.
D ✘	$\text{19}$
The answer should be somewhat smaller than $\text{19}$ because the standard deviation does not span the entire data set.

	$\left(x - \bar{x}\right)$ is squared because...
A	it prevents division by zero (prevents undefined answers).
B	it makes sure that we must use FOIL to square the binomial.
C	it changes negative distances into positive values.
D	mathematicians know students love complicated calculations.

A	$\text{16}$
B	$\text{0}$
C	$\text{317.95}$
D	$\text{4.01}$

A	$\text{12}$
B	$\text{209.9}$
C	$\text{0.56}$
D	$\text{3.03}$

A	$\text{419.6}$
B	$\text{10.29}$
C	$\text{0}$
D	$\text{40}$

A	$\text{27}$
B	$\text{29}$
C	$\text{30}$
D	$\text{35}$

	$\left(x - \bar{x}\right)$ is squared because...
A	it cancels out the subtraction effect.
B	it makes sure that we must use FOIL to square the binomial.
C	it changes negative distances into positive values.
D	exponents of $\text{2}$ make everything better.

	$\left(x - \bar{x}\right)$ is squared because...
A	it changes negative distances into positive values.
B	the formula must be quadratic to make a parabola.
C	it cancels out the subtraction effect.
D	exponents of $\text{2}$ make everything better.

A	Set 1, because it is the most evenly spread out.
B	Set 2, because it has the most values close to the middle.
C	Set 3, because it has the most values away from the middle
D	All three data sets would have the same standard deviation.

A	Set 1, because it has the most values away from the middle
B	Set 2, because it is the most evenly spread out.
C	Set 3, because it has the most values close to the middle.
D	All three data sets would have the same standard deviation.

A	which number is the most common.
B	how many values there are in a data set.
C	how high the numbers go.
D	how much a data set is spread out.

A	how far apart the minimum and maximum values are.
B	which number is the most common.
C	how much a data set is spread out.
D	how many values there are in a data set.

A	The data values are equal to each other.
B	The data set includes positive and negative values with zero in the middle.
C	The data values are all equal to zero.
D	It is impossible to know exactly what it means without the data.

A	$\sigma$ should increase because incomes will get further apart.
B	$\sigma$ should decrease because incomes will get closer together.
C	$\sigma$ should increase because more people will get jobs.
D	Standard deviation should not change (it will not increase or decrease).

A	$\sigma$ should increase because there will be more low CD4 count values.
B	$\sigma$ should decrease because there will be fewer low CD4 count values.
C	$\sigma$ should increase because the average CD4 count will go up.
D	Standard deviation should not change (it will not increase or decrease).

Measures of dispersion

1. Revision of range

Working with the range of a data set

Working with the range of a data set

Working with the range of a data set

Finding the range by solving an equation

Finding the range by solving an equation

Finding the range by solving an equation

Median, interquartile range, box & whisker diagram

Median, interquartile range, box & whisker diagram

Median, interquartile range, box & whisker diagram

Percentile and quartiles

Percentile and quartiles

Percentile and quartiles

Percentile and rank in a data set

Percentile and rank in a data set

Percentile and rank in a data set

Applying percentile to a data set

Applying percentile to a data set

Applying percentile to a data set

The range

The range

The range

Calculating the range of a data set

Calculating the range of a data set

Calculating the range of a data set

2. Variance & standard deviation

Using ¯xx¯\bar{x} and σσ\sigma to compare data sets

Using ¯xx¯\bar{x} and σσ\sigma to compare data sets

Using ¯xx¯\bar{x} and σσ\sigma to compare data sets

Standard deviation from the mean

Standard deviation from the mean

Standard deviation from the mean

Sigma and the mean

Sigma and the mean

Sigma and the mean

Matching the mean and standard deviation on a graph

Matching the mean and standard deviation on a graph

Matching the mean and standard deviation on a graph

Comparing how values are spread out

Comparing how values are spread out

Comparing how values are spread out

Mean and standard deviation

Mean and standard deviation

Mean and standard deviation

Standard deviation values

Standard deviation values

Standard deviation values

Roll the dice

Roll the dice

Roll the dice

How data values impact σσ\sigma

How data values impact σσ\sigma

How data values impact σσ\sigma

How do the data values impact σσ\sigma?

How do the data values impact σσ\sigma?

How do the data values impact σσ\sigma?

Calculating standard deviation

Calculating standard deviation

Calculating standard deviation

Standard deviation and the spread of data

Standard deviation and the spread of data

Standard deviation and the spread of data

What does standard deviation measure?

What does standard deviation measure?

What does standard deviation measure?

Estimating the standard deviation from a graph

Estimating the standard deviation from a graph

Estimating the standard deviation from a graph

Grouping values using sigma

Grouping values using sigma

Grouping values using sigma

Features of a data set graph

Features of a data set graph

Features of a data set graph

What does σ=0σ=0\sigma = 0 mean?

What does σ=0σ=0\sigma = 0 mean?

What does σ=0σ=0\sigma = 0 mean?

The meaning of standard deviation

The meaning of standard deviation

Using $\bar{x}$ and $\sigma$ to compare data sets

Using $\bar{x}$ and $\sigma$ to compare data sets

Using $\bar{x}$ and $\sigma$ to compare data sets

How data values impact $\sigma$

How data values impact $\sigma$

How data values impact $\sigma$

How do the data values impact $\sigma$ ?

How do the data values impact $\sigma$ ?

How do the data values impact $\sigma$ ?

What does $\sigma = 0$ mean?

What does $\sigma = 0$ mean?

What does $\sigma = 0$ mean?